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.
- Mathematics मा Limit भनेको फलनको मान हो। जुन मान given point मा कन्सन्ट्रेट भएको हुन्छ।
- गणितज्ञ Henkel(१८७१) का अनुसार आजभोलि हामीले कुरा गर्ने limit को परिभाषाको Euclid को Elements मा भएको बुक 10 को Proposotion 1 मा आधारित छ जसमा Euclid र Archimedes ले चर्चा गर्ने गरेको Method of exhaustion को प्रयोग भएको छ।
- Calculus मा continuity, derivatives and integrals जस्ता कुराहरुको चर्चा गर्न limit को अध्य्यन जरुरी छ।
- limit को परिभाषामा प्रयोग गर्ने ε-δ notation को चर्चा १९औ शताब्दीबाट Bernard Bolzano ले शुरु गरेको पाइन्छ, औपचारिक रुपमा 1821 मा Augustin-Louis Cauchy र पछि Karl Weierstrass ले limit को परिभाषा मा ε-δ notation को प्रयोग सुरु गरेको पाईन्छ।
- limit को ठिक तल प्रयोग गर्ने arrow चिन्हको प्रयोग २०औँ शताब्दीमा G. H. Hardy ले उनको किताब a course of pure mathematics बाट सुरु गरेको पाइन्छ ।
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 near to 1, then we find the value of f(x) near to 2
This is shown in a table below.
| 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 |
Tha table value shos that, 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
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
- Given a number a, we choose δ-neighbourhood of a, δ is positive AND can be small enough as we like such that |x - a|< δ
- Now, we will try to find, ε-neighbourhood of a given number L, ε is positive AND can be small enough as satisfied such that |f(x) - L|< ε
- If a small change in δ implies a small change ε, then the limit exists at a.
- If a small change in δ implies a LARGE change ε, or vice versa, then the limit does NOT exist at a.
Example: Limit exist at x=a
How to use the applet
- Click on the "Show δ-neighbourhood of 'a'" check box.
- 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
- Now, Click on the "Show ε-neighbourhood of 'L'" check box.
- 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 लागी |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- determine a number δ>0
- 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
- 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. - Once a δ is found, any smaller delta is acceptable, so there are an infinite number of possible δ's that we can choose.
- the function has limit at given x
Example: Limit Does not exist at x=0
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
- \( \displaystyle \lim_{x \to a^{-}} f(x) \) exists=LHS
- \( \displaystyle \lim_{x \to a^{+}} f(x) \) exists=RHS
- LHS=RHS
- In Figure (1), We see that the graph of f(x) has a hole at a. In fact, f(a) is undefined.[But, Limit exists at x=2]
- In Figure (2), f(a) is defined, but the function has a jump at a.[So, Limit does NOT exist at x=2]
- In Figure (3), f(a) is defined, but the function has a gap at a.[But, Limit exists at x=2]
| \( f(x)=\frac{x^2-1}{x-1}\) | \( \small {f(x)= \begin{cases} x+1 & \text{for } x ≤ 2 \\ x+2 & \text{for } x > 2 \end{cases}} \) | \( \small { f(x)= \begin{cases} x+1 & \text{for } x \ne 2 \\ 4 & \text{for } x = 2 \end{cases}} \) |
Example 1
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.
Example 2
Find the limit if exist for a function \( f(x)=\frac{1}{1-x}\) at x=1Indeterminate 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.
- \(\frac{0}{0}\)
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.
-
\(\frac{\infty}{\infty}\)
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? -
\(0.\infty\)
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? -
\(\infty-\infty\)
In general ∞ − (number) = ∞, but (number) − ∞ = −∞. So who wins? -
\(\infty^{0}\)
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? -
\(1^{\infty}\)
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? -
\(0^{0}\)
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?
Limit of algrabic function
- \( \displaystyle \lim_{x\to a}\frac{x^n-a^n}{x-a}=n{a^{n-1}}\)
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
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 of triangle \(OPM=\frac{1}{2} \sin \theta \cos \theta\)
त्रिभुजको क्षेत्रफल \( \frac{1}{2} \text{ base} \times \text {height} \) हुने भएकोले \(\triangle OPM=\frac{1}{2} \sin \theta \cos \theta\) हुन्छ ।Area of sector \(OPA=\frac{1}{2} \theta\)
वृतको चाँदक्षेत्रको क्षेत्रफल \( \frac{1}{2}r^2 \theta\) हुने भएकोले sector \(OPA=\frac{1}{2} \theta\) हुन्छ ।Area of triangle \(OBA=\frac{1}{2} \tan \theta\)
त्रिभुजको क्षेत्रफल \( \frac{1}{2} \text{ base} \times \text {height} \) 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.
More Theorems on Limit of trigonometric function
-
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 - \(\displaystyle \lim_{x \to 0} \sin x=0\)
- \(\displaystyle \lim_{x \to 0} \cos x=1\)
- \(\displaystyle \lim_{x \to 0}\frac{\tan x}{x}=1\)
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 \(f(x)=e^x\)
- Domain of \(f (x) = \{-\infty , \infty \}\)
- Range of \(f (x) = \{0, \infty \}\)
Graph of two exponential function \(e^x, e^{-x} \)
Theorem on Limit of exponential function
-
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. - 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.
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
- Product property: \( \log a (x.y) = \log_ax + \log a_y \)
- Quotient property: \( \log_ a (x/y) = \log_ax - \log_ay \)
- Power property: \( \log _ ax^nn = n \log _ax \)
- \( \log_ a a = 1, \log_ a 1 = 0 \)
- \( \log_ a m = \log_ a b \times \log_ b m \)
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
- \(\displaystyle \lim_{x\to 0}\frac{\log_e(1+x)}{x}=1 \)
- \(\displaystyle \lim_{x\to 0}\frac{\log_e(1-x)}{-x}=1 \)
- \(\displaystyle \lim_{x\to 0}\frac{a^x-1}{x}=\log a\)
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
No comments:
Post a Comment