Limit of a function

Table of Contents

1. Meaning of x→a
2. Intuitive Definition of Limit
3. Empirical Definition of Limit
4. Indeterminate Form
5. Limit of algrabic function
6. Limit of trigonometric function
7. Limit of Exponential function
8. Limit of logarithmic function
9. Limit (BCB-Revised Edition 2020, Exercise 2, Page 379)
10. Additional Question (Limit)[Page 392]

Introduction

In Mathematics, limit is defined as a value that a function approaches for the given point. It is always concerns about the behavior of a function at a particular point.

Meaning of \( x\to a \)

Consider a function
\(f(x)= \frac{x^2-1}{x-1}\)
We know that function is NOT defined at x=1. However, what happens to f(x) near the value x=1?

If we substitute small values for x, then we find that the value of f(x) is approximately 2 near at x=1

	x<1						x>1
x	0.5	0.6	0.7	0.8	0.9	1	1.1	1.2	1.3	1.4	1.5
f(x)	1.5	1.6	1.7	1.8	1.9	2	2.1	2.2	2.3	2.4	2.5

The closer that x gets to 1, the closer the value of the function f(x) to 2.
In such cases, we call it
f(x)=2 as x tends to 1

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Intuitive Definition of Limit

Let 𝑓(𝑥) be a function defined at all values in an open interval containing a, with the possible exception of "a" itself, and let L be a real number.
Then we say that \( \displaystyle \lim_{x \to a} f(x) =L\) if
for a given number \(\epsilon > 0\), there exists a number \(\delta > 0\)such that
|f(x)-L|<\(\epsilon\) whenever |x-a|<\(\delta\)

In this definition

1. Given a number L, we choose ε-neighbourhood of L, ε is positive AND can be small enough as we like such that |f(x) - L|< ε
2. Now, we will try to find, δ-neighbourhood of a, δ is positive AND can be small enough as satisfied such that |x - a|< δ
3. If a small change in ε implies a small change δ, then the limit exists at a.
4. If a small change in ε implies a LARGE change δ, then the limit does NOT exist at a.

How to use the applet

1. Click on the "Show δ-neighbourhood of 'a'" check box.
2. Given a number a, adjust δ-neighbourhood of a (drag the point a or the slider δ) , so that |x - a|< δ , where x is any point inside the δ-neighbourhood
3. Now, Click on the "Show ε-neighbourhood of 'L'" check box.
4. Try to find ε-neighbourhood of L (largest distance from L) , such that |f(x) - L|< ε , where f(x) is any point inside the ε-neighbourhood and the result is valid for all |x - a|< δ
   
   सबै |x - a|< δ को लगी |f(x) - L|< ε  हुने गरि ε-zone बानउन सकिन्छ भने limit exist हुन्छ ।
   
   Once a ε is found, any higher ε is always accepted.
   Once a δ is satisfied, any smaller δ is always accepted.

More Explanation

The intuitive definition says that

1. determine a number δ>0
2. take any x in the region, i.e. between a+δ and a−δ, then this x will be closer to a, that is |x-a|<δ
3. identify the point on the graph that our choice of x gives, then this point on the graph will lie in the intersection of the ε region. This means that this function value f(x) will be closer to L , that is |f(x)-L|<ε
   Means
   if we take any value of x in the δ region then the graph for those values of x will lie in the ε region.
4. Once a δ is found, any smaller delta is acceptable, so there are an infinite number of possible δ's that we can choose.
5. the function has limit at given x

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Empirical Definition of Limit

Let 𝑓(𝑥) be a function defined at all values in an open interval containing a, with the possible exception of "a" itself, and let L be a real number.
Then we say that \( \displaystyle \lim_{x \to a} f(x) =L\) if
all of the following three conditions hold

1. \( \displaystyle \lim_{x \to a^{-}} f(x) \) exists=LHS
2. \( \displaystyle \lim_{x \to a^{+}} f(x) \) exists=RHS
3. LHS=RHS

For Example

1. In Figure (1). We see that the graph of f(x) has a hole at a. In fact, f(a) is undefined.[Limit exists at x=2]
2. In Figure (2), f(a) is defined, but the function has a jump at a.[Limit does NOT exist at x=2]
3. In Figure (3), f(a) is defined, but the function has a gap at a.[Limit exists at x=2]

\( f(x)=\frac{x^2-1}{x-1}\)	\( \small {f(x)= \begin{cases} x+1 & \text{for } x ≤ 2 \\ x+2 & \text{for } x > 2 \end{cases}} \)	\( \small { f(x)= \begin{cases} x+1 & \text{for } x \ne 2 \\ 4 & \text{for } x = 2 \end{cases}} \)

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Indeterminate Form

The term "indeterminate" in mathematics refers to a situation where the value of an expression cannot be determined or uniquely identified based solely on its form or appearance.

1. \(\frac{0}{0}\)
   

   In the case of the expression "\(\frac{0}{0}\)" it is called indeterminate because it doesn't provide enough information to definitively determine the value of the expression.

   For example

   \(\frac{1}{1}=1\)	\(\frac{1}{0}=\infty\)	\(\frac{0}{1}=0\)
   \(\frac{2}{2}=1\)	\(\frac{2}{0}=\infty\)	\(\frac{0}{2}=0\)
   \(\frac{3}{3}=1\)	\(\frac{3}{0}=\infty\)	\(\frac{0}{3}=0\)
   …	…	…
   \(\frac{a}{a}=1\)	\(\frac{a}{0}=\infty\)	\(\frac{0}{a}=0\)
   \(\frac{0}{0}=1\)	\(\frac{0}{0}=\infty\)	\(\frac{0}{0}=0\)

   Here, \(\frac{0}{0}\) creates a situation where there is uncertainty about how the fraction \(\frac{0}{0}\) as a whole behaves. In other words, knowing that both the numerator and denominator are approaching zero doesn't immediately mean \(\frac{0}{0}\) will approach a specific finite value, approach infinity, or approach zero. The behavior of the fraction depends on the specific functions involved and how they approach zero.

2. \(\frac{\infty}{\infty}\)
   Usually \(\frac{\infty}{number}=\infty\) and \(\frac{number}{\infty}=0\). So the top pulls the limit up to infinity and the bottom tries to pull it down to 0. So who wins?
3. \(0.\infty\)
   Usually 0 · (number) = 0 and (number) · ∞ = ∞. So one piece tries to pull the limit down to zero, and the other tries to pull it up to ∞. Does one side win?
4. \(\infty-\infty\)
   In general ∞ − (number) = ∞, but (number) − ∞ = −∞. So who wins?
5. \(\infty^{0}\)
   In general ∞ raised to any positive power should be equal to ∞, ∞ raised to a negative power is 0, and anything raised to the zero should be equal to 1. So who wins?
6. \(1^{\infty}\)
   Usually 1 raised to any power is just equal to 1. But fractions raised to the ∞ goes to zero, and numbers larger than 1 raised to the ∞ should go off to ∞. So where does \(1^{\infty}\) go?
7. \(0^{0}\)
   In general zero raised to any positive power is just zero, but but anything raised to the zero should be equal to 1. So which is it?

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Limit of algrabic function

1. \( \displaystyle \lim_{x\to a}\frac{x^n-a^n}{x-a}=n{a^{n-1}}\)
   

   Solution
   We know that
   \( \frac{x^n-a^n}{x-a}=\frac{(x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+...+a^{n-1})}{x-a}\)
   or \( \frac{x^n-a^n}{x-a}=(x^{n-1}+ax^{n-2}+a^2x^{n-3}+...+a^{n-1})\)
   Thus, taking limit as \( x \to a\), we get
   \(\displaystyle \lim_{x\to a} \frac{x^n-a^n}{x-a}=\lim_{x\to a}(x^{n-1}+ax^{n-2}+a^2x^{n-3}+...+a^{n-1})\)
   or \(\displaystyle \lim_{x\to a} \frac{x^n-a^n}{x-a}=(a^{n-1}+a.a^{n-2}+a^2.a^{n-3}+...+a^{n-1})\)
   or \(\displaystyle \lim_{x\to a} \frac{x^n-a^n}{x-a}=n a^{n-1}\)
   This completes the proof

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Limit of trigonometric function

Trigonometry is branch of mathematics that deals about Triangle. The trigonometric ratio with reference to an angle x is called trigonometric function. For example,
f(x)= sinx

In this section we learn about two very specific but important trigonometric limits, and how to use them; and other tricks to find most other limits of trigonometric functions. The first involves the sine function, and the limit is
\(\displaystyle \lim_{\theta \to 0}\frac{\sin \theta }{\theta}=1\)

Here's a graph of \(f(x)=\frac{\sin x }{x}\), showing that it has a hole at x = 0. Our task in this section will be to prove that the limit from both sides of this function is 1.

Theorems on Limit of trigonometric function

Area if triangle \(OPM=\frac{1}{2} \sin \theta \cos \theta\)

त्रिभुजको क्षेत्रफल 1/2 * आधार * उचाई हुने भएकोले \(\triangle OPM=\frac{1}{2} \sin \theta \cos \theta\) हुन्छ ।

Area if sector \(OPB=\frac{1}{2} \theta\)

वृतको चाँदक्षेत्रको क्षेत्रफल 1/2 * अर्धव्यास ² * केन्द्रिय कोण हुने भएकोले sector \(OPB=\frac{1}{2} \theta\) हुन्छ ।

Area if triangle \(OBA=\frac{1}{2} \tan \theta\)

त्रिभुजको क्षेत्रफल 1/2 * आधार * उचाई हुने भएकोले \( \triangle OBA=\frac{1}{2} \tan \theta\) हुन्छ ।

In the figure above
Area of triangle OMP=\(\frac{1}{2} \sin \theta \cos \theta\)
Area of sector OAP=\(\frac{1}{2} \theta \)
Area of triangle OAB=\(\frac{1}{2} \tan \theta\)
Now
Area of triangle OMP \(\le\) Area of sector OAP \(\le\) Area of triangle OAB
or \(\frac{1}{2} \sin \theta \cos \theta \le \frac{1}{2} \theta \le \frac{1}{2} \tan \theta \)
or \(\sin \theta \cos \theta \le \theta \le \tan \theta \)
or \(\cos \theta \le \frac{\theta}{\sin \theta } \le \frac{1}{\cos \theta} \)
or \(\frac{1}{\cos \theta} \ge \frac{\sin \theta }{\theta} \ge \cos \theta \)
Taking limit as \( \theta \to 0\), we get
\( \displaystyle \lim_{\theta \to 0} \frac{1}{\cos \theta} \ge \lim_{\theta \to 0}\frac{\sin \theta }{\theta} \ge \lim_{\theta \to 0} \cos \theta \)
or \( \displaystyle \frac{1}{\cos 0} \ge \lim_{\theta \to 0}\frac{\sin \theta }{\theta} \ge \cos 0 \)
or \( \displaystyle \frac{1}{1} \ge \lim_{\theta \to 0}\frac{\sin \theta }{\theta} \ge 1 \)
or \( \displaystyle 1 \ge \lim_{\theta \to 0}\frac{\sin \theta }{\theta} \ge 1 \)
or \( \displaystyle \lim_{\theta \to 0}\frac{\sin \theta }{\theta}=1 \)
This completes the proof.

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More Theorems on Limit of trigonometric function

1. The another important limit involves the cosine function, specifically the function
   \(\displaystyle \lim_{\theta \to 0}\frac{\cos \theta -1}{\theta}=0\)

   Here's a graph of \(f(x)=\frac{\cos x-1}{x}\), showing that it has a hole at x = 0. Our task in this section will be to prove that the limit from both sides of this function is 0.

   Prove that \(\displaystyle \lim_{x \to 0}\frac{1- \cos x}{x}=0 \)
   Solution
   The limit is
   \(\displaystyle \lim_{x \to 0}\frac{1- \cos x}{x} \)
   or \(\displaystyle \lim_{x \to 0}\frac{1- \cos x}{x} \times \frac{1+ \cos x}{1+ \cos x} \)
   or \(\displaystyle \lim_{x \to 0}\frac{1- \cos^2 x}{x(1+ \cos x)} \)
   or \(\displaystyle \lim_{x \to 0}\frac{\sin^2 x}{x(1+ \cos x)} \)
   or \(\displaystyle \lim_{x \to 0}\frac{\sin x}{x} \times \lim_{x \to 0} \frac{\sin x}{1+ \cos x}\)
   or \( 1 \times \frac{0}{1+ 1}\)
   or 0
   This completes the proof

2. \(\displaystyle \lim_{x \to 0} \sin x=0\)
3. \(\displaystyle \lim_{x \to 0} \cos x=1\)
4. \(\displaystyle \lim_{x \to 0}\frac{\tan x}{x}=1\)

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Limit of Exponential function

A function of the form \(f (x) = a^x\) where base ‘a’ is constant (a>0) and the exponent ‘x’ is variable, is called exponential function.

For example,
\(f (x) = 2^x\)
is an exponential function.

Graph of two exponential function \(2^x, 2^{-x} \)

The great Swiss mathematician Leonhard Euler (1707-1783) has introduced the number e (e = 2.7182818284….). This value e is useful to define exponential function.
The function \(f(x)=e^x\) is called standard exponential function.
In this definition of \(f(x)=e^x\)

1. Domain of \(f (x) = \{-\infty , \infty \}\)
2. Range of \(f (x) = \{0, \infty \}\)

Graph of two exponential function \(e^x, e^{-x} \)

Theorem on Limit of exponential function

1. Prove that \(\displaystyle \lim_{x \to \infty} \left( 1+\frac{1}{x} \right )^x =e\)
   Solution
   We know that
   1. \(e=1+1!+\frac{1}{2!}+\frac{1}{3!}+...\)
   2. \( (1+x)^n=1+nx+\frac{n)n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+...\)
   Thus
   \(\left( 1+\frac{1}{x} \right )^x =1+ x \frac{1}{x} +\frac{x(x-1)}{2!} \left( \frac{1}{x}\right)^2+\frac{x(x-1)(x-2)}{3!} \left( \frac{1}{x}\right)^3 +...\)
   or \( \left( 1+\frac{1}{x} \right )^x =1+1 +\frac{1(1-\frac{1}{x})}{2!} +\frac{1(1-\frac{1}{x})(1-\frac{2}{x})}{3!}+...\)
   Taking limit as \( x \to \infty \), we get
   \( \displaystyle \lim_{x \to \infty} \left( 1+\frac{1}{x} \right )^x =1+1 + \lim_{x \to \infty}\frac{1(1-\frac{1}{x})}{2!} +\lim_{x \to \infty}\frac{1(1-\frac{1}{x})(1-\frac{2}{x})}{3!}+...\)
   or \( \displaystyle \lim_{x \to \infty} \left( 1+\frac{1}{x} \right )^x =1+1 + \frac{1(1-0)}{2!} +\frac{1(1-0)(1-0)}{3!}+...\)
   or \( \displaystyle \lim_{x \to \infty} \left( 1+\frac{1}{x} \right )^x =1+1 + \frac{1}{2!} +\frac{1}{3!}+...\)
   or \( \displaystyle \lim_{x \to \infty} \left( 1+\frac{1}{x} \right )^x =e\)
   This completes the Proof.

2. Prove that \(\displaystyle \lim_{x\to 0}\frac{e^x-1 }{x}=1 \)
   Solution
   \(\displaystyle \lim_{x\to 0}\frac{e^x-1 }{x} \)
   or \(\displaystyle \lim_{x\to 0}\frac{\left ( 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right ) -1 }{x} \)
   or \(\displaystyle \lim_{x\to 0}\frac{\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+... }{x} \)
   or \(\displaystyle \lim_{x\to 0} \frac{1}{1!}+\frac{x}{2!}+\frac{x^2}{3!}+... \)
   or \(\frac{1}{1!}+\frac{0}{2!}+\frac{0^2}{3!}+... \)
   or 1
   This completes the solution.

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Limit of logarithmic function

A logarithm function is an exponent of exponential function. For example,
if \( {a^x}=y\), then \(x={\log_a}y\).
In this definition
Log is the exponent, (or, exponent= Log)
if
\(3^2=9\) then \(2 = \log_3 9\)

In general, a function of the form \(f (x) = \log_e x\) called logarithmic function.
where

• Domain of f (x) = \( (0, \infty )\)
• Range of f (x) =\( (-\infty , \infty ) \)

Properties of logarithmic function

1. Product property: \( \log a (x.y) = \log_ax + \log a_y \)
2. Quotient property: \( \log_ a (x/y) = \log_ax - \log_ay \)
3. Power property: \( \log _ ax^nn = n \log _ax \)
4. \( \log_ a a = 1, \log_ a 1 = 0 \)
5. \( \log_ a m = \log_ a b \times \log_ b m \)

Graph of exponential and logarithem function

Log is the reflection of exponential function about y=x line, which is shown in a graph given below

Theorems on Limit of logarithmic function

1. \(\displaystyle \lim_{x\to 0}\frac{\log_e(1+x)}{x}=1 \)
2. \(\displaystyle \lim_{x\to 0}\frac{\log_e(1-x)}{-x}=1 \)
3. \(\displaystyle \lim_{x\to 0}\frac{a^x-1}{x}=\log a\)
   We know that
   \( \frac{d}{dx} a^x= \frac{d}{dx} e^{\log (a^x)} \)
   or \( \frac{d}{dx} a^x= \frac{d}{dx} e^{x \log a} \)
   or \( \frac{d}{dx} a^x= \log a .e^{x \log a} \)
   or \( \frac{d}{dx} a^x= \log a .e^{\log a^x} \)
   or \( \frac{d}{dx} a^x= \log a. a^x \)
   Thus, the limit is
   \(\displaystyle \lim_{x\to 0}\frac{a^x-1}{x}\)
   or \(\displaystyle \lim_{x\to 0}\frac{ \frac{d}{dx}(a^x-1)}{\frac{d}{dx} (x)} \)
   or \(\displaystyle \lim_{x\to 0}\frac{ \log a . a^x}{1} \)
   or \(\log a .a^0 \)
   or \(\log a \)
   This completes the proof

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Limit (BCB-Revised Edition 2020, Exercise 2, Page 379)

Evaluate the following.

1. \( \displaystyle \lim_{x \to 0} \frac{\sin ax}{x} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{\sin ax}{x} \)
   
   or \( \displaystyle \lim_{x \to 0} \frac{\sin ax}{ax} .a \)
   
   or \( \left ( \displaystyle \lim_{x \to 0} \frac{\sin ax}{ax} \right ) .a \)
   
   or \( a\)
   
   This completes the solution

2. \( \displaystyle \lim_{x \to 0} \frac{\tan bx}{x} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{\tan bx}{x} \)
   
   or \( \displaystyle \lim_{x \to 0} \frac{\tan bx}{bx} .b \)
   
   or \( \left ( \displaystyle \lim_{x \to 0} \frac{\tan bx}{x} \right ) .b \)
   
   or \( b\)
   
   This completes the solution

3. \( \displaystyle \lim_{x \to 0} \frac{\sin mx}{\sin nx} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{\sin mx}{\sin nx} \)
   
   or \( \displaystyle \lim_{x \to 0} \left ( \frac{ \frac{\sin mx}{mx}}{\frac{\sin nx}{nx}} \right ) \frac{mx}{nx} \)
   
   or \( \left ( \frac{ \displaystyle \lim_{x \to 0} \frac{\sin mx}{mx}}{ \displaystyle \lim_{x \to 0} \frac{\sin nx}{nx}} \right ) \frac{m}{n} \)
   
   or \( \left ( \displaystyle \frac{ 1}{ 1} \right ) \frac{m}{n} \)
   
   or \( \frac{m}{n} \)
   
   This completes the solution

4. \( \displaystyle \lim_{x \to 0} \frac{\tan ax}{\tan bx} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{\tan ax}{\tan bx}\)
   
   or \( \displaystyle \lim_{x \to 0} \left ( \frac{ \frac{\tan ax}{ax}}{\frac{\tan bx}{bx}} \right ) \frac{ax}{bx} \)
   
   or \( \left ( \frac{ \displaystyle \lim_{x \to 0} \frac{\tan ax}{ax}}{ \displaystyle \lim_{x \to 0} \frac{\tan bx}{bx}} \right ) \frac{a}{b} \)
   
   or \( \left ( \displaystyle \frac{ 1}{ 1} \right ) \frac{a}{b} \)
   
   or \( \frac{a}{b} \)
   
   This completes the solution

5. \( \displaystyle \lim_{x \to 0} \frac{\sin px}{\tan qx} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{\sin px}{\tan qx}\)
   
   or \( \displaystyle \lim_{x \to 0} \left ( \frac{ \frac{\sin px}{px}}{\frac{\tan qx}{qx}} \right ) \frac{px}{qx} \)
   
   or \( \left ( \frac{ \displaystyle \lim_{x \to 0} \frac{\sin px}{px}}{ \displaystyle \lim_{x \to 0} \frac{\tan qx}{qx}} \right ) \frac{p}{q} \)
   
   or \( \left ( \displaystyle \frac{ 1}{ 1} \right ) \frac{p}{q} \)
   
   or \( \frac{p}{q} \)
   
   This completes the solution

6. \( \displaystyle \lim_{x \to a} \frac{\sin (x-a)}{x^2-a^2} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to a} \frac{\sin (x-a)}{x^2-a^2}\)
   
   or \( \displaystyle \lim_{x \to a} \frac{\sin (x-a)}{x-a} \frac{1}{x+a}\)
   
   or \( \left ( \displaystyle \lim_{(x-a) \to 0} \frac{\sin (x-a)}{(x-a)} \right ) \left ( \displaystyle \lim_{x \to a} \frac{1}{x+a} \right )\)
   
   or \( \left ( 1 \right ) \left ( \frac{1}{a+a} \right ) \)
   
   or \( \frac{1}{2a} \)
   
   This completes the solution

7. \( \displaystyle \lim_{x \to p} \frac{x^2-p^2}{\tan (x-p)} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to p} \frac{x^2-p^2}{\tan (x-p)}\)
   
   or \( \displaystyle \lim_{x \to p} (x+p) \frac{(x-p)}{\tan (x-p)}\)
   
   or \( \displaystyle \lim_{x \to p} (x+p) \left ( \displaystyle \lim_{(x-p) \to 0} \frac{(x-p)}{\tan (x-p)} \right ) \)
   
   or \( (p+p) \left ( 1\right )\)
   
   or \(2p \)
   
   This completes the solution

8. \( \displaystyle \lim_{x \to 0} \frac{\sin ax. \cos bx}{\sin cx} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{\sin ax. \cos bx}{\sin cx} \)
   
   or \( \displaystyle \lim_{x \to 0} \left ( \frac{ \frac{\sin ax}{ax} (\cos bx) }{\frac{\sin cx}{cx} } \right ) \frac{(ax)}{(cx)}\)
   
   or \( \frac{ \left ( \displaystyle \lim_{x \to 0} \frac{\sin ax}{ax} \right ) \left ( \displaystyle \lim_{x \to 0} \cos bx \right ) }{ \left ( \displaystyle \lim_{x \to 0} \frac{\sin cx}{cx} \right ) } \frac{a}{c} \)
   
   or \( \frac{ \left ( 1 \right ) \left ( 1 \right ) }{ \left ( 1 \right ) } \frac{a}{c} \)
   
   or \( \frac{a}{c}\)
   
   This completes the solution

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9. \( \displaystyle \lim_{x \to 0} \frac{1-\cos x}{x^2} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{1-\cos x}{x^2} \)
   
   or \( \displaystyle \lim_{x \to 0} \frac{1-\cos x}{x^2} . \frac{1+\cos x}{1+\cos x}\)
   
   or \( \displaystyle \lim_{x \to 0} \frac{1-\cos^2 x}{x^2} . \frac{1}{1+\cos x}\)
   
   or \( \displaystyle \lim_{x \to 0} \frac{\sin^2 x}{x^2} . \frac{1}{1+\cos x}\)
   
   or \( \displaystyle \lim_{x \to 0} \left (\frac{\sin x}{x}\right )^2 . \frac{1}{1+\cos x}\)
   
   or \( \left ( \displaystyle \lim_{x \to 0} \frac{\sin x}{x}\right )^2 . \left ( \displaystyle \lim_{x \to 0} \frac{1}{1+\cos x} \right )\)
   
   or \( \left ( 1 \right )^2 . \left ( \frac{1}{1+1} \right )\)
   
   or \( \frac{1}{2}\)
   
   This completes the solution

10. \( \displaystyle \lim_{x \to 0} \frac{1-\cos 6x}{x^2} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{1-\cos 6x}{x^2} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{1-\cos 6x}{x^2} . \frac{1+\cos 6x}{1+\cos 6x}\)
    
    or \( \displaystyle \lim_{x \to 0} \frac{1-\cos^2 6x}{x^2} . \frac{1}{1+\cos 6x}\)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\sin^2 6x}{x^2} . \frac{1}{1+\cos 6x}\)
    
    or \( \displaystyle \lim_{x \to 0} \left (\frac{\sin 6x}{6x}\right )^2 .6^2. \frac{1}{1+\cos 6x}\)
    
    or \( \left ( \displaystyle \lim_{x \to 0} \frac{\sin 6x}{6x}\right )^2 .36. \left ( \displaystyle \lim_{x \to 0} \frac{1}{1+\cos 6x} \right )\)
    
    or \( \left ( 1 \right )^2 .36. \left ( \frac{1}{1+1} \right )\)
    
    or \( 18\)
    
    This completes the solution

11. \( \displaystyle \lim_{x \to 0} \frac{1-\cos 9x}{x^2} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{1-\cos 9x}{x^2} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{1-\cos 9x}{x^2} . \frac{1+\cos 9x}{1+\cos 9x}\)
    
    or \( \displaystyle \lim_{x \to 0} \frac{1-\cos^2 9x}{x^2} . \frac{1}{1+\cos 9x}\)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\sin^2 9x}{x^2} . \frac{1}{1+\cos 9x}\)
    
    or \( \displaystyle \lim_{x \to 0} \left (\frac{\sin 9x}{9x}\right )^2 .9^2. \frac{1}{1+\cos 9x}\)
    
    or \( \left ( \displaystyle \lim_{x \to 0} \frac{\sin 9x}{9x}\right )^2 .81. \left ( \displaystyle \lim_{x \to 0} \frac{1}{1+\cos 9x} \right )\)
    
    or \( \left ( 1 \right )^2 .81. \left ( \frac{1}{1+1} \right )\)
    
    or \( \frac{81}{2}\)
    
    This completes the solution

12. \( \displaystyle \lim_{x \to 0} \frac{\cos ax-\cos bx}{x^2} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{\cos ax-\cos bx}{x^2} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\cos ax-\cos bx}{x^2} . \frac{\cos ax+\cos bx}{\cos ax+\cos bx} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\cos^2 ax-\cos^2 bx}{x^2} . \frac{1}{\cos ax+\cos bx} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{(1-\sin^2 ax)-(1-\sin^2 bx)}{x^2} . \frac{1}{\cos ax+\cos bx} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\sin^2 bx-\sin^2 ax}{x^2} . \frac{1}{\cos ax+\cos bx} \)
    
    or \( \displaystyle \lim_{x \to 0} \left ( \frac{\sin^2 bx}{x^2} -\frac{\sin^2 ax}{x^2} \right ). \frac{1}{\cos ax+\cos bx} \)
    
    or \( \displaystyle \lim_{x \to 0} \left ( \left ( \frac{\sin bx}{bx}\right )^2 .b^2- \left ( \frac{\sin ax}{ax} \right )^2.a^2 \right ). \frac{1}{\cos ax+\cos bx} \)
    
    or \( \left ( \left ( \displaystyle \lim_{x \to 0} \frac{\sin bx}{bx}\right )^2 .b^2- \left ( \displaystyle \lim_{x \to 0} \frac{\sin ax}{ax} \right )^2.a^2 \right ). \left ( \displaystyle \lim_{x \to 0} \frac{1}{\cos ax+\cos bx} \right ) \)
    
    or \( \left ( 1)^2 .b^2- \left ( 1 \right )^2.a^2 \right ). \left ( \frac{1}{1+1} \right ) \)
    
    or \( \frac{b^2- a^2}{2} \)
    
    This completes the solution

13. \( \displaystyle \lim_{x \to 0} \frac{\sin ax-\sin bx}{x} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{\sin ax-\sin bx}{x} \)
    
    or \( \displaystyle \lim_{x \to 0} \left ( \frac{\sin ax}{x}-\frac{\sin bx}{x} \right ) \)
    
    or \( \left ( \displaystyle \lim_{x \to 0} \frac{\sin ax}{ax}. a- \displaystyle \lim_{x \to 0} \frac{\sin bx}{bx}.b \right ) \)
    
    or \( \left ( 1.a- 1.b \right ) \)
    
    or \( a-b \)
    
    This completes the solution

14. \( \displaystyle \lim_{x \to 0} \frac{1-\cos px}{1-\cos qx} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{1-\cos px}{1-\cos qx} \)
    
    or \( \displaystyle \lim_{x \to 0} (1-\cos px)\frac{1}{1-\cos qx} \)
    
    or \( \displaystyle \lim_{x \to 0}\left ( (1-\cos px). \frac{1+\cos px}{1+\cos px} \right ) \left ( \frac{1}{1-\cos qx} .\frac{1+\cos qx}{1+\cos qx} \right ) \)
    
    or \( \displaystyle \lim_{x \to 0}\left ( \frac{1-\cos^2 px}{1+\cos px} \right ) \left ( \frac{1+\cos qx}{1-\cos^2 qx} \right ) \)
    
    or \( \displaystyle \lim_{x \to 0}\left ( \frac{\sin^2 px}{1+\cos px} \right ) \left ( \frac{1+\cos qx}{\sin^2 qx} \right ) \)
    
    or \( \displaystyle \lim_{x \to 0}\left ( \frac{\sin^2 px}{\sin^2 qx} \right ) \left ( \frac{1+\cos qx}{1+\cos px} \right ) \)
    
    or \( \displaystyle \lim_{x \to 0}\left ( \frac{\frac{\sin px}{px} }{\frac{\sin qx}{q}} . \frac{px}{qx}\right )^2 \left ( \frac{1+\cos qx}{1+\cos px} \right ) \)
    
    or \( \left ( \frac{\displaystyle \lim_{x \to 0} \frac{\sin px}{px}}{\displaystyle \lim_{x \to 0} \frac{\sin qx}{q}} . \frac{p}{q}\right )^2 \left (\displaystyle \lim_{x \to 0} \frac{1+\cos qx}{1+\cos px} \right ) \)
    
    or \( \left (\frac{p}{q}\right )^2 \left (\frac{1+1}{1+1} \right ) \)
    
    or \( \frac{p^2}{q^2}\)
    
    This completes the solution

---

15. \( \displaystyle \lim_{x \to 0} \frac{\tan x-\sin x}{x^3} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{\tan x-\sin x}{x^3} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan x (1-\cos x)}{x^3} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan x}{x} \frac{(1-\cos x)}{x^2} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan x}{x} \frac{(1-\cos^2 x)}{x^2} \frac{1}{1+\cos x} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan x}{x} \frac{(\sin^2 x}{x^2} \frac{1}{1+\cos x} \)
    
    or \(\left ( \displaystyle \lim_{x \to 0} \frac{\tan x}{x} \right ) \left ( \displaystyle \lim_{x \to 0} \frac{\sin x}{x} \right )^2 \left ( \displaystyle \lim_{x \to 0} \frac{1}{1+\cos x} \right ) \)
    
    or \(1.1.\frac{1}{1+1} \)
    
    or \(\frac{1}{2} \)
    
    This completes the solution

16. \( \displaystyle \lim_{x \to 0} \frac{\tan 2x-\sin 2x}{x^3} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{\tan 2x-\sin 2x}{x^3} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan 2x (1-\cos 2x)}{x^3} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan 2x}{x} \frac{(1-\cos 2x)}{x^2} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan 2x}{x} \frac{(1-\cos^2 2x)}{x^2} \frac{1}{1+\cos 2x} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan 2x}{2x}.(2) \frac{\sin^2 2x}{(2x)^2} .(2^2) \frac{1}{1+\cos 2x} \)
    
    or \(\left ( \displaystyle \lim_{x \to 0} \frac{\tan 2x}{2x} .(2)\right ) \left ( \displaystyle \lim_{x \to 0} \frac{\sin 2x}{2x} .(2) \right )^2 \left ( \displaystyle \lim_{x \to 0} \frac{1}{1+\cos 2x} \right ) \)
    
    or \(1(2).1.(4)\frac{1}{1+1} \)
    
    or \(4 \)
    
    This completes the solution

17. \( \displaystyle \lim_{x \to \frac{\pi}{2}} (\sec x -\tan x) \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to \frac{\pi}{2}} (\sec x -\tan x) \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{2}} \sec x (1 -\sin x) \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sec x (1 -\sin^2 x)}{1+\sin x} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sec x \cos^2 x}{1+\sin x} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{1+\sin x} \)
    
    or \( \frac{\cos \frac{\pi}{2}}{1+\sin \frac{\pi}{2}} \)
    
    or \( \frac{0}{1+1} \)
    
    or \( 0 \)
    
    This completes the solution

18. \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{\sec ^2 x-2}{\tan x-1}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{\sec ^2 x-2}{\tan x-1} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{1+\tan^2 x-2}{\tan x-1} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{\tan^2 x-1}{\tan x-1} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} \tan x+1\)
    
    or \( \tan \frac{\pi}{4}+1\)
    
    or \( 1+1=2\)
    
    This completes the solution

19. \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{2- \csc ^2 x}{1 -\cot x}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{2- \csc ^2 x}{1 -\cot x} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{2- (1+\cot ^2 x)}{1 -\cot x} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{2- 1-\cot ^2 x}{1 -\cot x} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{1-\cot^2 x}{1 -\cot x} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} 1+\cot x\)
    
    or \( 1+\cot \frac{\pi}{4}\)
    
    or \( 1+1=2\)
    
    This completes the solution

20. \( \displaystyle \lim_{x \to y} \frac{\tan x -\tan y}{x-y}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to y} \frac{\tan x -\tan y}{x-y} \)
    
    or \( \displaystyle \lim_{x \to y} \frac{\tan (x-y). (1+\tan x \tan y)} {x-y} \)
    
    or \( \displaystyle \lim_{x \to y} \frac{\tan (x-y)}{x-y}. (1+\tan x \tan y) \)
    
    or \( \displaystyle \lim_{(x-y) \to 0} \frac{\tan (x-y)}{x-y}. \displaystyle \lim_{x \to y} (1+\tan x \tan y) \)
    
    or \( 1. (1+\tan x \tan x) \)
    
    or \( 1. (1+\tan^2 x ) \)
    
    or \( \sec^2 x \)
    
    This completes the solution

21. \( \displaystyle \lim_{x \to y} \frac{\sin x -\sin y}{x-y}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to y} \frac{\sin x -\sin y}{x-y} \)
    
    or \( \displaystyle \lim_{x \to y} \frac{2 \cos (\frac{x+y}{2}) .\sin (\frac{x-y}{2})}{x-y}\)
    
    or \( \displaystyle \lim_{x \to y} 2 \cos (\frac{x+y}{2}) \frac{\sin (\frac{x-y}{2})}{x-y}\)
    
    or \( \displaystyle \lim_{x \to y} 2 \cos (\frac{x+y}{2}) \frac{\sin (\frac{x-y}{2})}{(\frac{x-y}{2})} . \frac{1}{2}\)
    
    or \( \left ( \displaystyle \lim_{x \to y} \cos \frac{x+y}{2} \right ) \left ( \displaystyle \lim_{x \to y} \frac{\sin \frac{x-y}{2}}{\frac{x-y}{2}} \right ) \)
    
    or \( \cos \frac{y+y}{2} ( 1 ) \)
    
    or \( \cos y\)
    
    This completes the solution

22. \( \displaystyle \lim_{x \to y} \frac{\cos x -\cos y}{x-y}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to y} \frac{\cos x -\cos y}{x-y} \)
    
    or \( \displaystyle \lim_{x \to y} \frac{-2 \sin (\frac{x+y}{2}) .\sin (\frac{x-y}{2})}{x-y}\)
    
    or \( \displaystyle \lim_{x \to y} -2 \sin (\frac{x+y}{2}) \frac{\sin (\frac{x-y}{2})}{x-y}\)
    
    or \( \displaystyle \lim_{x \to y} -2 \sin (\frac{x+y}{2}) \frac{\sin (\frac{x-y}{2})}{(\frac{x-y}{2})} . \frac{1}{2}\)
    
    or \( \left ( \displaystyle \lim_{x \to y} -\sin \frac{x+y}{2} \right ) \left ( \displaystyle \lim_{x \to y} \frac{\sin \frac{x-y}{2}}{\frac{x-y}{2}} \right ) \)
    
    or \( -\sin \frac{y+y}{2} ( 1 ) \)
    
    or \( -\sin y\)
    
    This completes the solution

23. \( \displaystyle \lim_{x \to \theta} \frac{x \cot \theta -\theta \cot x}{x -\theta}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to \theta} \frac{x \cot \theta -\theta \cot x}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{x \cot \theta - \theta \cot \theta + \theta \cot \theta - \theta \cot x}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{(x- \theta) \cot \theta + \theta (\cot \theta - \cot x)}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{(x- \theta) \cot \theta} {x -\theta} + \displaystyle \lim_{x \to \theta} \frac{\theta (\cot \theta - \cot x)}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \cot \theta - \displaystyle \lim_{x \to \theta} \frac{\theta (\cot x -\cot \theta )}{x -\theta} \)
    
    or \( \cot \theta - \theta \displaystyle \lim_{x \to \theta} \frac{ - \frac{\sin(x-\theta)}{\sin x \sin \theta} }{x -\theta} \)
    
    or \( \cot \theta + \theta \displaystyle \lim_{x \to \theta} \frac{\sin(x-\theta)}{x -\theta} \frac{1}{\sin x \sin \theta} \)
    
    or \( \cot \theta + \theta \left ( \displaystyle \lim_{x-\theta \to 0} \frac{\sin(x-\theta)}{x -\theta} \right ) \left ( \displaystyle \lim_{x \to \theta} \frac{1}{\sin x \sin \theta} \right ) \)
    
    or \( \cot \theta + \theta .1. \frac{1}{\sin^2 \theta} \)
    
    or \( \cot \theta + \theta \csc^2 \theta \)
    
    This completes the solution

24. \( \displaystyle \lim_{x \to \theta} \frac{x \cos \theta -\theta \cos x}{x -\theta}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to \theta} \frac{x \cos \theta -\theta \cos x}{x -\theta}\)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{x \cos \theta -\theta \cos \theta+\theta \cos \theta -\theta \cos x}{x -\theta}\)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{(x- \theta) \cos \theta+ \theta ( \cos \theta - \cos x)}{x -\theta}\)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{(x- \theta) \cos \theta }{x -\theta} + \frac{ \theta (\cos \theta - \cos x)}{x -\theta}\)
    
    or \( \displaystyle \lim_{x \to \theta} \cos \theta + \theta \displaystyle \lim_{x \to \theta} \frac{ ( 2 \sin (\frac{x+\theta}{2}). \sin (\frac{x-\theta}{2}))}{x -\theta}\)
    
    or \( \cos \theta + \theta \left ( \displaystyle \lim_{x \to \theta} \sin (\frac{x+\theta}{2}) \right ) \left (\displaystyle \lim_{x-\theta \to 0} \frac{ \sin (\frac{x-\theta}{2})}{ (\frac{x-\theta}{2})} \right ) \)
    
    or \( \cos \theta + \theta \left ( \sin \theta \right ) \left (1 \right ) \)
    
    or \( \cos \theta + \theta \sin \theta \)
    
    This completes the solution

25. \( \displaystyle \lim_{x \to 1} \frac{1+ \cos \pi x}{\tan ^2 \pi x}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 1} \frac{1+ \cos \pi x}{\tan ^2 \pi x}\)
    
    or \( \displaystyle \lim_{x \to 1} \frac{1- \cos (\pi-\pi x)}{\tan ^2 (\pi-\pi x)} \)
    
    or \( \displaystyle \lim_{x \to 1} \frac{1- \cos^2 (\pi-\pi x)}{\tan ^2 (\pi-\pi x)} . \frac{1}{1+ \cos (\pi-\pi x)} \)
    
    or \( \displaystyle \lim_{x \to 1} \frac{\sin^2 (\pi-\pi x)}{\tan ^2 (\pi-\pi x)} . \frac{1}{1+ \cos (\pi-\pi x)} \)
    
    or \( \left ( \displaystyle \lim_{x-1 \to 0} \cos (\pi-\pi x) \right )^2 . \left ( \displaystyle \lim_{x \to 1} \frac{1}{1+ \cos (\pi-\pi x)} \right ) \)
    
    or \( \left ( 1 \right )^2 . \left ( \frac{1}{1+ 1} \right ) \)
    
    or \( \frac{1}{2} \)
    
    This completes the solution

26. \( \displaystyle \lim_{x \to \theta} \frac{x \tan \theta -\theta \tan x}{x -\theta}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to \theta} \frac{x \tan \theta -\theta \tan x}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{x \tan \theta - \theta \tan \theta + \theta \tan \theta - \theta \tan x}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{(x- \theta) \tan \theta + \theta (\tan \theta - \tan x)}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{(x- \theta) \tan \theta} {x -\theta} + \displaystyle \lim_{x \to \theta} \frac{\theta (\tan \theta - \tan x)}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \tan \theta - \displaystyle \lim_{x \to \theta} \frac{\theta (\tan x -\tan \theta )}{x -\theta} \)
    
    or \( \tan \theta - \theta \displaystyle \lim_{x \to \theta} \frac{ (\tan x -\tan \theta )}{x -\theta} \)
    
    or \( \tan \theta - \theta \displaystyle \lim_{x \to \theta} \frac{ \tan (x- \theta )(1+\tan x \tan \theta)}{x -\theta} \)
    
    or \( \tan \theta - \theta \displaystyle \lim_{x \to \theta} \frac{ \tan (x- \theta )}{x -\theta} \displaystyle \lim_{x \to \theta} (1+\tan x \tan \theta)\)
    
    or \( \tan \theta - \theta .1.(1+\tan^2 \theta)\)
    
    or \( \tan \theta - \theta \sec^2 \theta\)
    
    This completes the solution

27. \( \displaystyle \lim_{\theta \to \frac{\pi}{4}} \frac{\cos \theta -\sin \theta}{\theta -\frac{\pi}{4}}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{\theta \to \frac{\pi}{4}} \frac{\cos \theta -\sin \theta}{\theta -\frac{\pi}{4}} \)
    
    or \( \displaystyle \lim_{\theta \to \frac{\pi}{4}} \frac{ \sqrt{2} (\cos \theta \frac{1}{\sqrt{2}}-\sin \theta \frac{1}{\sqrt{2}}) } {\theta -\frac{\pi}{4}} \)
    
    or \( \sqrt{2} \displaystyle \lim_{\theta \to \frac{\pi}{4}} \frac{\cos \theta \sin \frac{\pi}{4} -\sin \theta \cos \frac{\pi}{4}}{\theta -\frac{\pi}{4}}\)
    
    or \( - \sqrt{2} \displaystyle \lim_{\theta \to \frac{\pi}{4}} \frac{\sin \theta \cos \frac{\pi}{4} -\cos \theta \sin \frac{\pi}{4}}{\theta -\frac{\pi}{4}}\)
    
    or \( - \sqrt{2} \displaystyle \lim_{\theta \to \frac{\pi}{4}} \frac{\sin (\theta -\frac{\pi}{4}) }{\theta -\frac{\pi}{4}}\)
    
    or \( - \sqrt{2} .1\)
    
    or \( - \sqrt{2}\)
    
    This completes the solution

28. \( \displaystyle \lim_{x \to c} \frac{\sqrt{x}-\sqrt{c}}{\sin x-\sin c}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to c} \frac{\sqrt{x}-\sqrt{c}}{\sin x-\sin c} \)
    
    or \( \displaystyle \lim_{x \to c} \frac{x-c}{(\sin x-\sin c)(\sqrt{x}+\sqrt{c})} \)
    
    or \( \displaystyle \lim_{x \to c} \frac{x-c}{\sin x-\sin c} \frac{1}{\sqrt{x}+\sqrt{c}} \)
    
    or \( \displaystyle \lim_{x \to c} \frac{x-c}{2 \cos (\frac{x+c}{2}) \sin (\frac{x-c}{2}) } \frac{1}{\sqrt{x}+\sqrt{c}} \)
    
    or \( \displaystyle \lim_{x \to c} \frac{\frac{x-c}{2}}{ \sin (\frac{x-c}{2}) } \frac{1}{\cos (\frac{x+c}{2}) (\sqrt{x}+\sqrt{c})} \)
    
    or \( \displaystyle \lim_{x \to c} \frac{\frac{x-c}{2}}{ \sin (\frac{x-c}{2}) } \displaystyle \lim_{x \to c} \frac{1}{\cos (\frac{x+c}{2}) (\sqrt{x}+\sqrt{c})} \)
    
    or \( 1. \frac{1}{\cos c (\sqrt{c}+\sqrt{c})} \)
    
    or \( \frac{\sec c}{2\sqrt{c}} \)
    
    This completes the solution

---

29. Find the limits of
    1. \( \displaystyle \lim_{x \to 0} \frac{e^{6x}-1}{x}\)
       

       Solution 👉 Click Here

       The limit takes the indeterminate form \(\frac{0}{0}\), we proceed further calculation
       Given that
       \( \displaystyle \lim_{x \to 0} \frac{e^{6x}-1}{x} \)
       
       or \( \left ( \displaystyle \lim_{x \to 0} \frac{e^{6x}-1}{6x} \right ).6\) \( \because \displaystyle \lim_{x \to 0} \frac{e^{x}-1}{x}=1\)
       
       or \( 1.6\)
       
       or \( 6\)
       
       This completes the solution

    2. \( \displaystyle \lim_{x \to 0} \frac{e^{2x}-1}{x. 2^{x+1}}\)
       

       Solution 👉 Click Here

       The limit takes the indeterminate form \(\frac{0}{0}\), we proceed further calculation
       Given that
       \( \displaystyle \lim_{x \to 0} \frac{e^{2x}-1}{x. 2^{x+1}} \)
       
       or \( \displaystyle \lim_{x \to 0} \frac{e^{2x}-1}{x} . \frac{1}{2^{x+1}}\)
       
       or \( \displaystyle \lim_{x \to 0} \frac{e^{2x}-1}{2x} . \frac{1}{2^{x}}\)
       
       or \( \left ( \displaystyle \lim_{x \to 0} \frac{e^{2x}-1}{2x} \right ) . \left (\displaystyle \lim_{x \to 0} \frac{1}{2^{x}} \right ) \)
       
       or \( \left ( 1 \right ) . \left (\frac{1}{1} \right ) \) \( \because \displaystyle \lim_{x \to 0} \frac{e^{x}-1}{x}=1\)
       
       or \( 1\)
       
       This completes the solution

    3. \( \displaystyle \lim_{x \to 0} \frac{e^{ax}-e^{bx}}{x}\)
       

       Solution 👉 Click Here

       The limit takes the indeterminate form \(\frac{0}{0}\), we proceed further calculation
       Given that
       \( \displaystyle \lim_{x \to 0} \frac{e^{ax}-e^{bx}}{x} \)
       
       or\( \displaystyle \lim_{x \to 0} \frac{e^{ax}-1-e^{bx}+1}{x} \)
       
       or\( \displaystyle \lim_{x \to 0} \frac{(e^{ax}-1)-(e^{bx}-1)}{x} \)
       
       or \( \displaystyle \lim_{x \to 0} \frac{e^{ax}-1}{x}- \displaystyle \lim_{x \to 0} \frac{e^{bx}-1}{x} \)
       
       or \( \displaystyle \lim_{x \to 0} \frac{e^{ax}}{ax} .a - \displaystyle \lim_{x \to 0} \frac{e^{bx}}{bx} .b \)
       
       or \( 1 .a - 1 .b \) \( \because \displaystyle \lim_{x \to 0} \frac{e^{x}-1}{x}=1\)
       
       or \( a - b \)
       
       This completes the solution

    ---

    4. \( \displaystyle \lim_{x \to 0} \frac{a^x+b^x-2}{x}\)
       

       Solution 👉 Click Here

       The limit takes the indeterminate form \(\frac{0}{0}\), we proceed further calculation
       Given that
       \( \displaystyle \lim_{x \to 0} \frac{a^x+b^x-2}{x} \)
       
       or\( \displaystyle \lim_{x \to 0} \frac{a^x-1+b^x-1}{x} \)
       
       or\( \displaystyle \lim_{x \to 0} \frac{a^x-1}{x}+ \lim_{x \to 0} \frac{b^x-1}{x} \)
       
       or\( \log a+ \log b \) \( \because \displaystyle \lim_{x \to 0} \frac{a^x-1}{x}=\log a\)
       
       or\( \log (ab)\)
       
       This completes the solution

30. Evaluate the limits of
    1. \( \displaystyle \lim_{x \to 2} \frac{x-2}{\log (x-1)}\)
       

       Solution 👉 Click Here

       The limit takes the indeterminate form \(\frac{0}{0}\), we proceed further calculation
       Given that
       \( \displaystyle \lim_{x \to 2} \frac{x-2}{\log (x-1)} \)
       
       or\( \displaystyle \lim_{x \to 2} \frac{x-2}{\log [1+(x-2)]} \)
       
       or\( \displaystyle \lim_{x \to 2} \frac{1}{\frac{\log [1+(x-2)]}{(x-2)}} \)
       
       or\( \frac{1}{ \displaystyle \lim_{x-2 \to 0} \frac{\log [1+(x-2)]}{(x-2)}} \)
       
       or\( \frac{1}{ 1} \) \( \because \displaystyle \lim_{x \to 0} \frac{\log (1+x)}{x}=1\)
       
       or\( 1 \)
       
       This completes the solution

    2. \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\log \left ( x- \frac{\pi}{2} +1\right )}\)
       

       Solution 👉 Click Here

       The limit takes the indeterminate form \(\frac{0}{0}\), we proceed further calculation
       Given that
       \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\log \left ( x- \frac{\pi}{2} +1\right )} \)
       
       or\( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sin (\frac{\pi}{2}-x)}{\log \left [ 1+(x- \frac{\pi}{2}) \right ]} \)
       
       or\( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{- \sin (x-\frac{\pi}{2})}{\log \left [ 1+(x- \frac{\pi}{2}) \right ]} \)
       
       or\( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{- \sin (x-\frac{\pi}{2})}{(x-\frac{\pi}{2})} . \frac{x-\frac{\pi}{2}}{\log \left [ 1+(x- \frac{\pi}{2}) \right ]} \)
       
       or\( - \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{ \sin (x-\frac{\pi}{2})}{(x-\frac{\pi}{2})} . \frac{1}{ \displaystyle \lim_{x- \frac{\pi}{2} \to 0} \frac {\log \left [ 1+(x- \frac{\pi}{2}) \right ] }{ (x-\frac{\pi}{2})} } \)
       
       or\( - 1 . 1 \)
       
       or\( - 1 \)
       
       This completes the solution

---

---

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Additional Question (Limit)[Page 392]

1. Prove that \( \displaystyle \lim_{x \to \frac{2}{3}} \frac{2}{2-3x}\) does NOT exist.

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to \frac{2}{3}} \frac{2}{2-3x}\)
   
   At \(x= \frac{2}{3}\), we compute the following
   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to \frac{2}{3}^-} f(x)= \lim_{h \to 0^-} \frac{2}{2-3(\frac{2}{3}-h)}=\lim_{h \to 0^-}\frac{2}{2-2+h}=\lim_{h \to 0^-}\frac{2}{h}=\) ∞
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to \frac{2}{3}^+} f(x)= \lim_{h \to 0^+} \frac{2}{2-3(\frac{2}{3}+h)}=\lim_{h \to 0^+}\frac{2}{2-2-h}=\lim_{h \to 0^+}\frac{2}{-h}=\) -∞
   
   Since,
   LHL≠RHL
   
   The limit does NOT exist at \(x= \frac{2}{3}\)

   
   This completes the solution
2. Do the following function define for the value x=1?
   1. \( f(x)=\frac{x-1}{x+2}\)
      
      

      Solution 👉 Click Here

      Yes, the function \( f(x)=\frac{x-1}{x+2}\) is defined for the value x=1, because when we put x=1 to the function f(x), then
      
      \( f(x)=\frac{x-1}{x+2}\)
      
      = \( f(x)=\frac{1-1}{1+2}\)
      
      = \( f(x)=\frac{0}{3}\)
      
      = \( f(x)=0\)
      
      
   2. \( f(x)=\frac{x^3+1}{x-1}\)

      Solution 👉 Click Here

      No, the function \( f(x)=\frac{x^3+1}{x-1}\) is NOT defined for the value x=1, because when we put x=1 to the function f(x), then
      
      \( f(x)=\frac{x^3+1}{x-1}\)
      
      = \( f(x)=\frac{1^3+1}{1-1}\)
      
      = \( f(x)=\frac{2}{0}\)
      
      = \( f(x)=\infty\)
      
      
3. What do you mean by the left hand limit and right hand limit of a function? What is the condition for the limit of a function to exist at a point?
   Prove that \(\displaystyle \lim_{x \to 0}|x|=0\) but \(\displaystyle \lim_{x \to 0} \frac{|x|}{x} \) does not exist.

   Solution 👉 Click Here

   Left-hand limit

   The left-hand limit of a function f(x) at a particular point x=a is used to determine the behavior of the function f(x) as it approaches a from the left. Left-Hand Limit (LHL): is defined as
   \( LHL=\displaystyle \lim_{x \to a^-} f(x)\)
   
   This limit represents the behavior of the function f(x) as x gets closer to a while staying to the left of a.

   Right-hand limit

   The right-hand limit of a function f(x) at a particular point x=a is used to determine the behavior of the function f(x) as it approaches a from the right. Right-Hand Limit (RHL): is defined as
   \( RHL=\displaystyle \lim_{x \to a^+} f(x)\)
   
   This limit represents the behavior of the function f(x) as x gets closer to a while staying to the right of a.

   Condition for the limit of a function to exist at a point

   let f(x) be a function, then condition for the limit of a function f(x) to exist at a point a is that
   \( LHL=RHL\)
   
   or \( \displaystyle \lim_{x \to a^-} f(x)=\lim_{x \to a^+} f(x)\)
   
   This limit represents the behavior of the function f(x) as x gets closer to the point a, from both sides:left and right.

   \(\displaystyle \lim_{x \to 0}|x|=0\)

   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to 0^-} |x|= \lim_{x \to 0^-} -x = \) 0
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to 0^+} |x|= \lim_{x \to 0^+} x = \) 0
   
   Since,
   LHL=RHL
   
   The limit exist at x=0, and the limit value is 0

   \(\displaystyle \lim_{x \to 0} \frac{|x|}{x} \) does NOT exist

   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to 0^-} \frac{|x|}{x}= \lim_{x \to 0^-} \frac{-x}{x} = \) -1
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to 0^+} \frac{|x|}{x}= \lim_{x \to 0^+} \frac{x}{x} = \) 1
   
   Since,
   LHL≠RHL
   
   The limit does NOT exist at x=0

4. Distinguish between limit and value of a function.
   It is given that \(f(x)=\frac{ax+b}{x+1},\displaystyle \lim_{x \to 0} f(x)=2\) and \( \displaystyle \lim_{x \to \infty} f(x)=1\). Prove that \(f(-2)=0\)

   Solution 👉 Click Here

   Distinguish between limit and value

   	Limit of a Function	Value of a Function
   Behaviour at a	represents the behavior of the function as it gets closer and closer to a particular point (typically denoted as "a")	The value of a function represents the actual output or result of the function when a specific input is provided.
   Symbolically	the limit of a function f(x) as x approaches a is denoted as \( \displaystyle \lim_{x \to a} f(x) \)	the value of a function f(x) at a particular point a is denoted as f(a)
   Definition at a	The limit may not necessarily be equal to the function's actual value at that point. It describes what happens to the function's values as x gets arbitrarily close to a.	It provides information about the function's behavior at a specific point.
   	Limits are used to analyze the behavior of functions near singular/critical points, to determine continuity, derivatives and integrals.	Functional value are used to compute the value at a

   Problem Solving

   It is given that \(f(x)=\frac{ax+b}{x+1}\)
   
   Also given that
   \(\displaystyle \lim_{x \to 0} f(x)=2\)
   
   or\(\displaystyle \lim_{x \to 0} \frac{ax+b}{x+1}=2\)
   
   or\( \frac{a.0+b}{0+1}=2\)
   
   or\( b=2\)
   
   Also given that
   \( \displaystyle \lim_{x \to \infty} f(x)=1\)
   
   or\( \displaystyle \lim_{x \to \infty} \frac{ax+b}{x+1}=1\)
   
   or\( \displaystyle \lim_{x \to \infty} \frac{a+\frac{b}{x}}{1+\frac{1}{x}}=1\)
   
   or\( \frac{a+\frac{b}{\infty}}{1+\frac{1}{\infty}}=1\)
   
   or\( \frac{a+0}{1+0}=1\)
   
   or\( a=1\)
   
   Now
   \(f(x)=\frac{ax+b}{x+1}\)
   
   or\(f(-2)=\frac{a.(-2)+b}{(-2)+1}\)
   
   or\(f(-2)=\frac{1.(-2)+2}{(-2)+1}\)
   
   or\(f(-2)=0\)
   
   This completes the proof.

5. Define limit of a function at a point. It is given that \(f(x)=\frac{x+6}{cx-d},\displaystyle \lim_{x \to 0} f(x)=-6 \) and \( \displaystyle \lim_{x \to \infty} f(x)=\frac{1}{3}\).
   Prove that \(f(13)=\frac{1}{2}\)

   Solution 👉 Click Here

   Limit of a function

   Let 𝑓(𝑥) be a function defined at all values in an open interval containing a, with the possible exception of "a" itself, and let L be a real number.
   Then we say that \( \displaystyle \lim_{x \to a} f(x) =L\) if for a given number \(\epsilon > 0\), there exists a number \(\delta > 0\)such that
   |f(x)-L|<\(\epsilon\) whenever |x-a|<\(\delta\)

   Empirical Definition of Limit

   Let 𝑓(𝑥) be a function defined at all values in an open interval containing a, with the possible exception of "a" itself, and let L be a real number.
   Then we say that \( \displaystyle \lim_{x \to a} f(x) =L\) if all of the following three conditions hold

   1. \( \displaystyle \lim_{x \to a^{-}} f(x) \) exists=LHS
   2. \( \displaystyle \lim_{x \to a^{+}} f(x) \) exists=RHS
   3. LHS=RHS

   Problem Solving

   It is given that \(f(x)=\frac{x+6}{cx-d}\)
   
   Also given that
   \(\displaystyle \lim_{x \to 0} f(x)=-6\)
   
   or\(\displaystyle \lim_{x \to 0} \frac{x+6}{cx-d}=-6\)
   
   or\( \frac{0+6}{c.0-d}=-6\)
   
   or\( d=1\)
   
   Also given that
   \( \displaystyle \lim_{x \to \infty} f(x)=\frac{1}{3}\)
   
   or\( \displaystyle \lim_{x \to \infty} \frac{x+6}{cx-d}=\frac{1}{3}\)
   
   or\( \displaystyle \lim_{x \to \infty} \frac{1+\frac{6}{x}}{c-\frac{d}{x}}=\frac{1}{3}\)
   
   or\( \frac{1+0}{c-0}=\frac{1}{3}\)
   
   or\( \frac{1}{c}=\frac{1}{3}\)
   
   or\( c=3\)
   
   Now
   \(f(x)=\frac{x+6}{cx-d}\)
   
   or\(f(13)=\frac{13+6}{c.13-d}\)
   
   or\(f(13)=\frac{13+6}{3.13-1}\)
   
   or\(f(13)=\frac{1}{2}\)
   
   This completes the proof.

6. What do you mean by an indeterminate form? State their different forms. Evaluate the following limit \(\displaystyle \lim_{x \to \infty} \sqrt{x} (\sqrt{x}-\sqrt{x-a})\)

   Solution 👉 Click Here

   Indeterminate form

   The term "indeterminate" refers to a situation where the value of an expression cannot be uniquely determined based solely on its appearance.

   The different form of such indeterminates are as follows.

   1. \(\frac{0}{0}\)
      

      The expression "\(\frac{0}{0}\)" is called indeterminate because it doesn't provide enough information to uniquely determine its value

      For example

      \(\frac{1}{1}=1\)	\(\frac{1}{0}=\infty\)	\(\frac{0}{1}=0\)
      \(\frac{2}{2}=1\)	\(\frac{2}{0}=\infty\)	\(\frac{0}{2}=0\)
      \(\frac{3}{3}=1\)	\(\frac{3}{0}=\infty\)	\(\frac{0}{3}=0\)
      …	…	…
      \(\frac{a}{a}=1\)	\(\frac{a}{0}=\infty\)	\(\frac{0}{a}=0\)
      \(\frac{0}{0}=1\)	\(\frac{0}{0}=\infty\)	\(\frac{0}{0}=0\)

      Here, \(\frac{0}{0}\) creates a situation where there is uncertainty about how the fraction \(\frac{0}{0}\) as a whole behaves. In other words, knowing that both the numerator and denominator are approaching zero doesn't immediately mean \(\frac{0}{0}\) will approach a specific finite value, approach infinity, or approach zero. So, it is indeterminate

   2. \(\frac{\infty}{\infty}\)
      Usually \(\frac{\infty}{number}=\infty\) and \(\frac{number}{\infty}=0\). So the top pulls the limit up to infinity and the bottom tries to pull it down to 0. So who wins?
   3. \(0.\infty\)
      Usually 0 · (number) = 0 and (number) · ∞ = ∞. So one piece tries to pull the limit down to zero, and the other tries to pull it up to ∞. Does one side win?
   4. \(\infty-\infty\)
      In general ∞ − (number) = ∞, but (number) − ∞ = −∞. So who wins?
   5. \(\infty^{0}\)
      In general ∞ raised to any positive power should be equal to ∞, ∞ raised to a negative power is 0, and anything raised to the zero should be equal to 1. So who wins?
   6. \(1^{\infty}\)
      Usually 1 raised to any power is just equal to 1. But fractions raised to the ∞ goes to zero, and numbers larger than 1 raised to the ∞ should go off to ∞. So where does \(1^{\infty}\) go?
   7. \(0^{0}\)
      In general zero raised to any positive power is just zero, but but anything raised to the zero should be equal to 1. So which is it?

   Problem Solving

   \(\displaystyle \lim_{x \to \infty} \sqrt{x} (\sqrt{x}-\sqrt{x-a})\)	= \(\displaystyle \lim_{x \to \infty} \sqrt{x} (\sqrt{x}-\sqrt{x-a}) \frac{(\sqrt{x}+\sqrt{x-a})}{(\sqrt{x}+\sqrt{x-a})}\)
   	=\(\displaystyle \lim_{x \to \infty} \sqrt{x} (x-x+a) \frac{1}{(\sqrt{x}+\sqrt{x-a})}\) =\(\displaystyle \lim_{x \to \infty} \frac{a\sqrt{x}}{(\sqrt{x}+\sqrt{x-a})}\) =\(\displaystyle \lim_{x \to \infty} \frac{a.1}{(1+\sqrt{1-\frac{a}{x}})}\) =\( \frac{a}{(1+\sqrt{1-0})}\) =\( \frac{a}{(1+1)}\) =\( \frac{a}{2}\)

7. Let \(f:R \to R \) be defined by \(f(x)=\begin{cases} x & \text{if x is an integer} \\ 0 & \text{if x is not an integer} \\ \end{cases}\)
   Find \( \displaystyle \lim_{x \to 1} f(x)\). Is it same as \(f(1)\)

   Solution 👉 Click Here

   Here

   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to 1^-} x= \lim_{x \to 1^-} 0 = \) 0
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to 1^+} x= \lim_{x \to 1^+} 0 = \) 0
   
   Since,
   LHL=RHL
   
   The limit exist at x=1, and the limit value is 0

   Next,
   The functional value at x=1, is
   \( f(x)= x =\) 1
   
   So, The functional value and limit value is NOT same.
8. Prove that
   1. \(\displaystyle \lim_{x \to 3} \left ( \frac{1}{x-3}-\frac{9}{x^3-3x^2}\right ) =\frac{2}{3}\)

      Solution 👉 Click Here

      Solution
      
      \(\displaystyle \lim_{x \to 3} \left ( \frac{1}{x-3}-\frac{9}{x^3-3x^2}\right ) \)
      
      =\(\displaystyle \lim_{x \to 3} \left ( \frac{1}{x-3}-\frac{9}{x^2(x-3)}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \frac{1}{x-3} \left ( 1-\frac{9}{x^2}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \frac{1}{x-3} \left ( \frac{x^2-9}{x^2}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \left ( \frac{x+3}{x^2}\right )\)
      
      = \(\frac{3+3}{3^2}\)
      
      = \(\frac{2}{3}\)
      
      

   2. \(\displaystyle \lim_{x \to 3} \left ( \frac{x^2+9}{x^2-9}-\frac{3}{x-3}\right )=\frac{1}{2}\)

      Solution 👉 Click Here

      Solution
      
      \(\displaystyle \lim_{x \to 3} \left ( \frac{x^2+9}{x^2-9}-\frac{3}{x-3}\right )\)
      
      =\(\displaystyle \lim_{x \to 3} \left ( \frac{x^2+9}{(x+3)(x-3)}-\frac{3}{x-3}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \left ( \frac{x^2+9-3(x+3)}{(x+3)(x-3)}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \left ( \frac{x^2-3x}{(x+3)(x-3)}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \left ( \frac{x(x-3)}{(x+3)(x-3)}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \left ( \frac{x}{(x+3)}\right )\)
      
      = \(\frac{3}{3+3}\)
      
      = \(\frac{1}{2}\)
      
      

9. Evaluate
   1. \(\displaystyle \lim_{x \to 2} \frac{x^{-3}-2^{-3}}{x-2}\)

      Solution 👉 Click Here

      Solution
      
      \(\displaystyle \lim_{x \to 2} \frac{x^{-3}-2^{-3}}{x-2}\)
      
      =\( (-3) (2)^{-3-1}\) Using formula \(\displaystyle \lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}\), taking n=-3, a=2
      
      =\( (-3) (2)^{-4}\)
      
      =\( \frac{-3}{2^4}\)
      
      =\( \frac{-3}{16}\)
      
      

   2. \(\displaystyle \lim_{x \to \infty} \frac{(2x-1)^6(3x-1)^4}{(2x+1)^{10}}\)

      Solution 👉 Click Here

      Solution
      
      \(\displaystyle \lim_{x \to \infty} \frac{(2x-1)^6(3x-1)^4}{(2x+1)^{10}}\)
      
      =\(\displaystyle \lim_{x \to \infty} \frac{(2-\frac{1}{x})^6(3-\frac{1}{x})^4}{(2+\frac{1}{x})^{10}}\) Dividing by \(x^{10}\)
      
      =\(\frac{(2-0)^6(3-0)^4}{(2+0)^{10}}\)
      
      =\(\frac{3^4}{2^4}\)
      
      =\( \frac{81}{16}\)
      
      

   3. \(\displaystyle \lim_{x \to 0} \frac{(1+x)^6-1}{(1+x)^2-1}\)

      Solution 👉 Click Here

      Solution
      
      \(\displaystyle \lim_{x \to 0} \frac{(1+x)^6-1^6}{(1+x)^2-1^2}\)
      
      = \(\displaystyle \lim_{(1+x) \to 1} \frac{\frac{(1+x)^6-1^6}{(1+x)-1} }{\frac{(1+x)^2-1^2}{(1+x)-1} } \)
      
      =\( \frac{6.1^{6-1}}{2.1^{2-1}} \) Using formula \(\displaystyle \lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}\)
      
      =\(\frac{6}{2}\)
      
      =\(3\)
      
      

   4. \(\displaystyle \lim_{x \to a} \frac{(x+2)^{\frac{5}{2}} -(a+2)^{\frac{5}{2}} }{x-a}\)

      Solution 👉 Click Here

      Solution
      
      \(\displaystyle \lim_{x \to a} \frac{(x+2)^{\frac{5}{2}} -(a+2)^{\frac{5}{2}} }{x-a}\)
      
      = \(\displaystyle \lim_{(x+2) \to (a+2)} \frac{(x+2)^{\frac{5}{2}} -(a+2)^{\frac{5}{2}} }{(x+2)-(a+2)}\)
      
      =\( \frac{5}{2}.(a+2)^{\frac{5}{2}-1} \) Using formula \(\displaystyle \lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}\) taking \(n=\frac{5}{2}\), a=a+2
      
      =\( \frac{5}{2}.(a+2)^{\frac{3}{2}} \)
      
      

10. If \( \displaystyle \lim_{x \to a} \frac{x^3-a^3}{x-a}=27\), Find all possible values of a.

    Solution 👉 Click Here

    Solution
    
    \( \displaystyle \lim_{x \to a} \frac{x^3-a^3}{x-a}=27\)
    
    =\( 3.(a)^{3-1}=27 \) Using formula \(\displaystyle \lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}\) taking \(n=3\)
    
    =\( 3a^2=27 \)
    
    =\( a^2=9 \)
    
    =\( a=\pm 3 \)
    
    

11. Find the limiting values of
    1. \( \displaystyle \lim_{x \to 0} \frac{\sin x^0}{x}\)

       Solution 👉 Click Here

       Solution
       
       \( \displaystyle \lim_{x \to 0} \frac{\sin x^0}{x}\)
       
       = \( \displaystyle \lim_{x \to 0} \frac{\sin (\frac{\pi x}{180})}{x}\)
       
       = \( \displaystyle \lim_{x \to 0} \frac{\sin (\frac{\pi x}{180})}{\frac{\pi x}{180}} \frac{\pi}{180}\)
       
       =\( 1.\frac{\pi}{180} \)
       
       =\( \frac{\pi}{180} \)
       
       

    2. \( \displaystyle \lim_{x \to 0} \frac{1-\cos 4 x}{1-\cos 6x}\)

       Solution 👉 Click Here

       Solution
       
       \( \displaystyle \lim_{x \to 0} \frac{1-\cos 4 x}{1-\cos 6x}\)
       
       = \( \displaystyle \lim_{x \to 0} \frac{1-\cos^2 4 x}{1+\cos 4 x}.\frac{1+\cos 6 x}{1-\cos^2 6x}\)
       
       = \( \displaystyle \lim_{x \to 0} \frac{\sin^2 4 x}{1+\cos 4 x}.\frac{1+\cos 6 x}{\sin^2 6x}\)
       
       = \( \displaystyle \lim_{x \to 0} \frac{\sin^2 4 x}{(4x)^2} . \frac{(6x)^2}{\sin^2 6 x} .\frac{1+\cos 6 x}{1+\cos 4 x} .\frac{(4x)^2}{(6x)^2}\)
       
       = \( \displaystyle \lim_{x \to 0} \left (\frac{\sin 4 x}{4x} \right )^2 \left (\frac{6x}{\sin 6 x} \right )^2 \left ( \frac{1+\cos 6 x}{1+\cos 4 x} \right ) \left (\frac{4^2}{6^2} \right ) \)
       
       = \( \left (1 \right )^2 \left (1 \right )^2 \left ( \frac{1+0}{1+0} \right ) \left (\frac{4}{9} \right ) \)
       
       = \( \frac{4}{9} \)
       
       

    3. \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2}-x}\)

       Solution 👉 Click Here

       Solution
       
       \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2}-x}\)
       
       = \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sin (\frac{\pi}{2}-x)} {(\frac{\pi}{2}-x)}\)
       
       = \(1\)
       
       

    4. \( \displaystyle \lim_{x \to 0} \frac{\tan 2 x-x}{3x-\sin x}\)

       Solution 👉 Click Here

       \( \displaystyle \lim_{x \to 0} \frac{\tan 2 x-x}{3x-\sin x}\)	=\( \displaystyle \lim_{x \to 0} \frac{\frac{\tan 2 x}{2}-1}{3-\frac{\sin x}{x}}\)
       	=\( \frac{\displaystyle \lim_{x \to 0} \frac{\tan 2 x}{2x}.2-1}{3-\displaystyle \lim_{x \to 0} \frac{\sin x}{x}}\) =\(\frac{1.2-1}{3-1}\) =\(\frac{1}{2}\)

    5. \( \displaystyle \lim_{x \to \pi} \frac{1-\sin (\frac{x}{2})}{(\pi-x)^2}\)

       Solution 👉 Click Here

       Solution
       
       \( \displaystyle \lim_{x \to \pi} \frac{1-\sin (\frac{x}{2})}{(\pi-x)^2}\)
       
       = \( \displaystyle \lim_{x \to \pi} \frac{1-\cos (\frac{\pi}{2} -\frac{x}{2})}{(\pi-x)^2}\)
       
       = \( \displaystyle \lim_{x \to \pi} \frac{1-\cos (\frac{\pi-x}{2})}{(\pi-x)^2}\)
       
       = \( \displaystyle \lim_{x \to \pi} \frac{1-\cos^2 (\frac{\pi-x}{2})}{(\pi-x)^2} . \frac{1}{1+\cos (\frac{\pi-x}{2})}\)
       
       = \( \displaystyle \lim_{x \to \pi} \frac{\sin^2 (\frac{\pi-x}{2})}{(\pi-x)^2}. \frac{1}{1+\cos (\frac{\pi-x}{2})}\)
       
       = \( \displaystyle \lim_{x \to \pi} \left (\frac{\sin^2 (\frac{\pi-x}{2})}{(\frac{\pi-x}{2})} \right )^2. (\frac{1}{2})^2 .\frac{1}{1+\cos (\frac{\pi-x}{2})}\)
       
       = \( (1 )^2. \frac{1}{4} .\frac{1}{1+1}\)
       
       = \(\frac{1}{8}\)
       
       

    6. \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1+\cos 2x}{(\pi-2x)^2}\)

       Solution 👉 Click Here

       Solution
       
       \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1+\cos 2x}{(\pi-2x)^2}\)
       
       = \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1-\cos (\pi-2x)}{(\pi-2x)^2}\)
       
       = \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1-\cos^2 (\pi-2x)}{(\pi-2x)^2}. \frac{1}{1+\cos (\pi-2x)} \)
       
       = \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sin^2 (\pi-2x)}{(\pi-2x)^2}. \frac{1}{1+\cos (\pi-2x)} \)
       
       = \( \displaystyle \lim_{x \to \frac{\pi}{2}} \left ( \frac{\sin (\pi-2x)}{(\pi-2x)} \right )^2. \lim_{x \to \frac{\pi}{2}} \frac{1}{1+\cos (\pi-2x)} \)
       
       = \( (1)^2 \frac{1}{1+1} \)
       
       = \( \frac{1}{2} \)
       
       

    7. \( \displaystyle \lim_{x \to 0} \sin (\frac{1}{x})\)

       Solution 👉 Click Here

       Let \(\frac{1}{x}=u\) then
       
       \( \displaystyle \lim_{x \to 0} \sin (\frac{1}{x}) = \displaystyle \lim_{u \to \infty} \sin u \)
       
       The left hand limit is
       
       As \( x \to 0^-\) then \(\displaystyle \lim_{x \to 0^-} \frac{1}{x} =\lim_{h \to 0^-} \frac{1}{0-h}=-\infty\)
       
       Thus
       
       \(\displaystyle \lim_{u \to -\infty} \sin (u)=\sin (-\infty)=- \sin \infty \)
       
       The right hand limit is
       
       As \( x \to 0^+\) then \(\displaystyle \lim_{x \to 0^+} \frac{1}{x} =\lim_{h \to 0^+} \frac{1}{0+h}=\infty\)
       
       Thus
       
       \(\displaystyle \lim_{u \to \infty} \sin (u)= \sin (\infty) \)
       
       We see that, function \( \sin u\) goes up to infinity from the right. From the left,it goes down to negative infinity.
       So, the limit of a function \( \sin (\frac{1}{x}) \) does NOT exist at x=0.

       For more justification

       1. given the function \( f(x)=\sin \frac{1}{x} \)
       2. take \( x=\frac{1}{\frac{\pi}{2} (2n+1)};n \in Z \) then \( \frac{1}{x}=\frac{\pi}{2} (2n+1);n \in Z \) and \( \sin (\frac{1}{x})=\sin[\frac{\pi}{2} (2n+1)] \)
       3. then value of \(f(x)= \sin[\frac{\pi}{2} (2n+1)] \) are alternatively to 1 and -1 as x approaches to 0
       4. so, \( f(x)=\sin \frac{1}{x} \) has oscillating discontinuity at 0

       Figure 1

       Figure 2

       Figure 3

       Figure 4

       Figure 5
    8. \( \displaystyle \lim_{x \to 0} x \sin (\frac{1}{x})\)

       Solution 👉 Click Here

       Let \(\frac{1}{x}=u\) then
       
       \( \displaystyle \lim_{x \to 0} x \sin (\frac{1}{x}) = \displaystyle \lim_{u \to \infty} \frac{1}{u} \sin u \)
       
       The left hand limit is
       
       As \( x \to 0^-\) then \(\displaystyle \lim_{x \to 0^-} \frac{1}{x} =\lim_{h \to 0^-} \frac{1}{0-h}=-\infty\)
       
       Thus
       
       \(\displaystyle \lim_{u \to -\infty} \frac{1}{u} \sin (u)= \frac{1}{\infty }\sin (-\infty)=0. (\sin \infty) =0\)
       
       The right hand limit is
       
       As \( x \to 0^+\) then \(\displaystyle \lim_{x \to 0^+} \frac{1}{x} =\lim_{h \to 0^+} \frac{1}{0+h}=\infty\)
       
       Thus
       
       \(\displaystyle \lim_{u \to \infty} \frac{1}{u} \sin (u)= \frac{1}{\infty} \sin (\infty) =0. \sin (\infty)=0\)
       
       We see that, function \( \frac{1}{u} \sin (u)\) goes up to 0 from both side.
       So, the limit of a function \( x \sin (\frac{1}{x}) \) exist at x=0, and the limit value is 0.

       For more justification
    9. \( \displaystyle \lim_{x \to a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}}\)

       Solution 👉 Click Here

       Solution
       
       \( \displaystyle \lim_{x \to a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}}\)
       
       = \( \displaystyle \lim_{x \to a} \frac{2 \cos (\frac{x+a}{2})\sin (\frac{x-a}{2})}{x-a}. (\sqrt{x}+\sqrt{a})\)
       
       = \( \left ( \displaystyle \lim_{x \to a} \frac{\sin (\frac{x-a}{2}) }{(\frac{x-a}{2}) } \right ) . \left ( \displaystyle \lim_{x \to a} \cos (\frac{x+a}{2}) .(\sqrt{x}+\sqrt{a}) \right )\)
       
       = \( (1) .\cos a .(2\sqrt{a}) \)
       
       = \( 2\sqrt{a}\cos a \)
       
       

    10. \( \displaystyle \lim_{x \to a} (a-x)\tan (\frac{\pi x}{2a})\)

        Solution 👉 Click Here

        Solution
        \( \displaystyle \lim_{x \to a} (a-x)\tan (\frac{\pi x}{2a})\)
        
        = \( \displaystyle \lim_{x \to a} (a-x)\tan (\frac{\pi x}{2a})\)
        
        = \( \displaystyle \lim_{x \to a} (a-x)\cot (\frac{\pi }{2}-\frac{\pi x}{2a})\)
        
        = \( \displaystyle \lim_{x \to a} (a-x)\cot (\frac{\pi(a-x) }{2a})\)
        
        = \( \displaystyle \lim_{x \to a} \frac{a-x}{\tan (\frac{\pi(a-x) }{2a})}\)
        
        = \( \displaystyle \lim_{x \to a} \frac{(\frac{\pi(a-x) }{2a})} {\tan (\frac{\pi(a-x) }{2a}) } . (\frac{2a}{\pi}) \)
        
        = \( \frac{2a}{\pi}\)
        
        

        Solution Method 2

        Let a-x=h, then \(x \to a\) implies \(h \to 0\) and \( \frac{x}{a}=1+\frac{h}{a}\)

        Solution
        
        \( \displaystyle \lim_{x \to a} (a-x)\tan (\frac{\pi x}{2a})\)
        
        = \( \displaystyle \lim_{h \to 0} (h)\tan \frac{\pi}{2} (1+\frac{h}{a})\)
        
        = \( \displaystyle \lim_{h \to 0} (h)\tan (\frac{\pi}{2}+\frac{\pi h}{2 a})\)
        
        = \( \displaystyle \lim_{h \to 0} (h)\cot (\frac{\pi h}{2 a})\)
        
        = \( \displaystyle \lim_{h \to 0} \frac{h}{\tan (\frac{\pi h}{2 a})}\)
        = \( \displaystyle \lim_{h \to 0} \frac{(\frac{\pi h}{2 a})}{\tan (\frac{\pi h}{2 a})}. (\frac{2a}{\pi})\)
        
        = \( \frac{2a}{\pi}\)
        
        

    11. \( \displaystyle \lim_{y \to 0} \frac{(x+y)\sec (x+y)-x \sec x}{y}\)

        Solution 👉 Click Here

        \( \displaystyle \lim_{y \to 0} \frac{(x+y)\sec (x+y)-x \sec x}{y}\)

        =\( \displaystyle \lim_{y \to 0} \frac{x\sec (x+y)+y\sec (x+y)-x \sec x}{y}\)

        =\( \displaystyle \lim_{y \to 0} \frac{x\sec (x+y)-x \sec x}{y} + \lim_{y \to 0} \frac{y\sec (x+y)}{y} \) =\( \displaystyle x \lim_{y \to 0} \frac{\sec (x+y)- \sec x}{y} +\lim_{y \to 0} \sec (x+y) \) =\( \displaystyle x \lim_{y \to 0} \frac{\cos x-\cos (x+y)}{y\cos x \cos (x+y)} +\sec x \) =\( \displaystyle x \lim_{y \to 0} \frac{-2 \sin (\frac{x+x+y}{2}) \sin (\frac{x-x-y}{2})}{y\cos x \cos (x+y)} +\sec x \) =\( \displaystyle x \lim_{y \to 0} \frac{2 \sin (x+\frac{y}{2}) \sin (\frac{y}{2})}{y\cos x \cos (x+y)} +\sec x \) =\( \displaystyle x \lim_{y \to 0} \frac{ \sin (x+\frac{y}{2}) }{\cos x \cos (x+y)}.\lim_{y \to 0} \frac{ \sin (\frac{y}{2})}{\frac{y}{2}} +\sec x \) =\( x \frac{ \sin (x+0) }{\cos x \cos (x+0)}.1 +\sec x \) =\( x \frac{ \sin x }{\cos x \cos x} +\sec x \) =\( x \tan x \sec x +\sec x \)

12. 
    
    1. A function is defined as \(f(x)=\begin{cases} 3x^2+2 & \text { if } x<1 \\ 2x+3 & \text { if } x \ge 1 \end{cases}\).
       Find \(\displaystyle \lim_{x \to 1}f(x) \)

       Solution 👉 Click Here

       Solution
       
       

       The left hand limit is
       \( LHL=\displaystyle \lim_{x \to 1^-} 3x^2+2= 3.1^2+2= \) 5
       
       The right hand limit is
       \(RHL= \displaystyle \lim_{x \to 1^+} 2x+3= 2.1+3= \) 5
       
       Since,
       LHL=RHL
       
       The limit exist at x=1, and the limit value is 5
    2. A function is defined as \(f(x)=\begin{cases} 3+2x & \text { if } -\frac{3}{2} \le x < 0\\ 3-2x & \text { if } 0 \le x < \frac{3}{2} \\ -3-2x & \text { if } x \ge \frac{3}{2} \end{cases}\).
       Find \(\displaystyle \lim_{x \to 0}f(x) \) and \(\displaystyle \lim_{x \to \frac{3}{2}}f(x) \) if they exist.

       Solution 👉 Click Here

       Solution \(\displaystyle \lim_{x \to 0}f(x) \)
       
       

       The left hand limit is
       \( LHL=\displaystyle \lim_{x \to 0^-} 3+2x= 3+2.0 = \) 3
       
       The right hand limit is
       \(RHL= \displaystyle \lim_{x \to 0^+} 3-2x=3-2.0 = \) 3
       
       Since,
       LHL=RHL
       
       The limit exist at x=0, and the limit value is 3

       Solution \(\displaystyle \lim_{x \to 3/2}f(x) \)
       
       

       The left hand limit is
       \( LHL=\displaystyle \lim_{x \to \frac{3}{2}^-} 3-2x= 3-3= \) 0
       
       The right hand limit is
       \(RHL= \displaystyle \lim_{x \to \frac{3}{2}^+} -3-2x=-3-3 = \) -6
       
       Since,
       LHL≠RHL
       
       The limit does NOT exist at x=3/2

13. 
    
    1. \(\displaystyle \lim_{x \to 0} \frac{e^{px}-1}{e^{qx}-1} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to 0} \frac{e^{px}-1}{e^{qx}-1} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^{px}-1}{px} . \frac{qx}{e^{qx}-1} . \frac{px}{qx}\)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^{px}-1}{px} . \lim_{x \to 0} \frac{qx}{e^{qx}-1} . \frac{p}{q}\)
       
       = \(1.1 . \frac{p}{q}\) Using formula \(\displaystyle \lim_{x \to 0} \frac{e^x-1}{x}=1\)
       
       = \(\frac{p}{q}\)
       
       

    2. \(\displaystyle \lim_{x \to 0} \frac{e^x-e^{-x}-x}{x} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to 0} \frac{e^x-e^{-x}-x}{x} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^x-1-e^{-x}+1-x}{x} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^x-1}{x}-\frac{e^{-x}-1}{x}-\frac{x}{x} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^x-1}{x}+\lim_{x \to 0} \frac{e^{-x}-1}{-x}-1\)
       
       = \(1+1-1\) Using formula \(\displaystyle \lim_{x \to 0} \frac{e^x-1}{x}=1\)
       
       = \(1+1-1\)
       
       

    3. \(\displaystyle \lim_{x \to 0} \frac{a^x-1}{b^x-1} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to 0} \frac{a^x-1}{b^x-1} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{a^x-1}{x} . \frac{x}{b^x-1} \) Using formula \(\displaystyle \lim_{x \to 0} \frac{a^x-1}{x}=\log a\)
       
       = \(\log a. \frac{1}{\log b}\)
       
       = \(\log (a-b)\)
       
       

14. 
    
    1. \(\displaystyle \lim_{x \to 0} \frac{2^x-1}{\sin x} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to 0} \frac{2^x-1}{\sin x} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{2^x-1}{x}. \lim_{x \to 0} \frac{x}{\sin x} \) Using formula \(\displaystyle \lim_{x \to 0} \frac{a^x-1}{x}=\log a\), taking a=2
       
       = \(\log 2. 1\)
       
       = \(\log 2\)
       
       

    2. \(\displaystyle \lim_{x \to 0} \frac{e^{\sin x} -\sin x -1}{x} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to 0} \frac{e^{\sin x} -\sin x -1}{x} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^{\sin x}-1}{x} -\lim_{x \to 0} \frac{\sin x}{x} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^{\sin x}-1}{\sin x} .\lim_{x \to 0} \frac{\sin x}{x} - \lim_{x \to 0} \frac{\sin x}{x} \)
       
       = \( 1 .1 - 1 \) Using formula \(\displaystyle \lim_{x \to 0} \frac{e^x-1}{x} =1\)
       
       = \(0\)
       
       

    3. \(\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{e^{\cos x} -1}{\frac{\pi}{2}-x} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{e^{\cos x} -1}{\frac{\pi}{2}-x} \)
       
       = \(\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{e^{\sin (\frac{\pi}{2}-x)} -1}{\frac{\pi}{2}-x} \)
       
       = \(\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{e^{\sin (\frac{\pi}{2}-x)} -1}{\sin (\frac{\pi}{2}-x)} . \lim_{x \to 0} \frac{\sin (\frac{\pi}{2}-x)}{\frac{\pi}{2}-x} \)
       
       = \(1 .1 \) Using formula \(\displaystyle \lim_{x \to 0} \frac{e^x-1}{x} =1\)
       
       = \(1 \)
       
       

    4. \(\displaystyle \lim_{x \to e} \frac{\log x-1}{x-e} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to e} \frac{\log x-1}{x-e} \)
       
       = \(\displaystyle \lim_{x \to e} \frac{1}{e}.\frac{\log x-\log e}{\frac{x}{e}-1} \)
       
       = \(\frac{1}{e} . \displaystyle \lim_{x \to e} \frac{\log (\frac{x}{e})}{\frac{x}{e}-1} \)
       
       = \(\frac{1}{e} . \displaystyle \lim_{x \to e} \frac{\log (1+[\frac{x}{e}-1])}{[\frac{x}{e}-1]} \)
       
       = \(\frac{1}{e} \) Using formula \(\displaystyle \lim_{x \to 0} \frac{\log (1+x)}{x} =1\)