Introduction
Many functions have the property that their graphs can be traced with a pencil without lifting the pencil from the page. Such functions are called continuous.
Intuitively, a function is continuous at a particular point if there is no break in its graph at that point.
let’s consider some examples that fail to meet our intuitive notion of what it means to be continuous at a point.
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| \( f(x)=\frac{x^2-1}{x-1}\) | \( \small {f(x)= \begin{cases} x+1 & \text{for } x ≤ 2 \\ x+2 & \text{for } x > 2 \end{cases}} \) | \( \small { f(x)= \begin{cases} x+1 & \text{for } x \ne 2 \\ 4 & \text{for } x = 2 \end{cases}} \) |
- In Figure (1). We see that the graph of f(x) has a hole at a. In fact, f(a) is undefined. So f(x) is NOT contnuous
- In Figure (2), f(a) is defined, but the function has a jump at a. So f(x) is NOT contnuous.
- In Figure (3), f(a) is defined, but the function has a gap at a. So f(x) is NOT contnuous.
Definition
A function f(x) is defined to be continuous at a if the following three conditions holds
- f(a) is defined
- \( \displaystyle \lim_{x\to a^{-}}f(x)\) exist = LHL
- \( \displaystyle \lim_{x\to a^{+}}f(x) \) exist = RHL
- LHL = RHL = f(a)
Examples
-
A function \( f(x) = x^2 + 1 \) is continuous at 2, since
\( \displaystyle \lim_{x \rightarrow 2} f(x) =5 =f(2)\)
-
A function \( f(x) = \sqrt{4-x^2} \) is NOT continuous at 3
since f(3) is not defined.
-
A function \( f(x)=\frac{1}{x-2}\)
is not continuous at 2 because f(2) is not defined.
NOTES
- The constant function, f (x) = c, \( \forall x \in R\) is continuous on R.
- The identity function, f (x) = x, \( \forall x \in R\) is continuous on R.
- The function f (x) = x n, \( x \in N\) is continuous on R.
- The function f (x) = sinx is continuous.
Example 1
Test the continuity of a function \( f(x)= \begin{cases} -x^2 &\text { for } x \ne 2 \\ 0 &\text {for } x=2 \end{cases} \) at x=2.
Solution
The solution is as follows
At \( x=2^{-}\), the left hand limit is
\( \displaystyle \lim_{x \to 2^{-}} f(x)= \lim_{x \to 2^{-} } -x^2 =-4\)
LHL
At \( x=2^{+}\), the right hand limit is
\( \displaystyle \lim_{x \to 2^{+}} f(x)= \lim_{x \to 2^{+}} = -x^2 =-4\)
RHL
At x=2, the value of the function is
f(2) = 0
Functional Value
Since, limiting value and functional value are not equal, f(x) is not continuous at x=2.
Example 2
Test the continuity of a function \( f(x)= \begin{cases} -x^2 +2 &\text { for } x \ne 2 \\ 1 &\text { for } x=2 \end{cases} \) at x=2.
Solution
The solution is as follows
At \( x=2^{-}\), the left hand limit is
\( \displaystyle \lim_{x \to 2^{-}} f(x)= \lim_{x \to 2^{-} } -x^2+2 =-2\)
LHL
At \( x=2^{+}\), the right hand limit is
\( \displaystyle \lim_{x \to 2^{+}} f(x)= \lim_{x \to 2^{+}} = -x^2+2 =-2\)
RHL
At x=2, the value of the function is
f(2) = 1
Functional Value
Since, limiting value and functional value are not equal, f(x) is not continuous at x=2.
Example 3
Test the continuity of a function \(f(x)=\frac{|x|}{x}\) for \(x\ne 0\) at x=0.
Solution
The solution is
At x = 0, the value of f(0) is not defined.
Also, the left hand limit is -1, whereas the right hand limit is 1
So, the function is not continuous at x=0.
Example 4
Is\( f(x)=\frac{x+4}{x-3}\) continuous at x =1? At x =3?
Solution
The solution is as follows
At x=1, the limit value is
\( \displaystyle \lim_{x \to 1} f(x)= \lim_{x \to 1}\frac{x+4}{x-3}=\lim_{x \to 1} \frac{1+4}{1-3} =\frac{5}{-2}\)
This is Limit Value
At x=1, the value of the function is
\( \displaystyle \lim_{x \to 1} f(x)= \lim_{x \to 1} \frac{1+4}{1-3} =\frac{5}{-2}\)
This is Functional Value
Since, limiting value and functional value are equal, f(x) is continuous at x=1.
Next,
At x=3, the limit value is
\( \displaystyle \lim_{x \to 3} f(x)= \lim_{x \to 3} \frac{x+4}{x-3} =\frac{7}{0}\)=undefined
Since, the limiting value is undefined, f(x) is NOT continuous at x=3.
Exercise: BCB-Revised Edition 2020, Exercise 15.3, Page 389
- Test the continuity of discontinuity of the following function by calculating the left hand limits, the right-hand limits and the values of the functions at points specified.
- \( f(x)=x^2\) at x=4
Solution 👉 Click Here
Given that
\( f(x)=x^2 \)
At x=4, we compute the following
The functional value is
\(f(x)= x^2 = 4^2=\) 16
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 4^-} f(x)= \lim_{x \to 4^-} x^2 = 4^2=\) 16
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 4^+} f(x)= \lim_{x \to 4^+} x^2 = 4^2=\) 16
Since,
LHL=RHL=f(x)
The function is continuous at x=4
This completes the solution
- \( f(x)=2-3x^2\) at x=0
Solution 👉 Click Here
Given that
\( f(x)=2-3x^2 \)
At x=0, we compute the following
The functional value is
\(f(x)= 2-3x^2= 2-3.0^2=\) 2
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 0^-} f(x)= \lim_{x \to 0} 2-3x^2= 2-3.0^2=\) 2
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 0^+} f(x)= \lim_{x \to 0} 2-3x^2= 2-3.0^2=\) 2
Since,
LHL=RHL=f(x)
The function is continuous at x=0
This completes the solution
- \( f(x)=3x^2-2x+4\) at x=1
Solution 👉 Click Here
Given that
\( f(x)=3x^2-2x+4 \)
At x=1, we compute the following
The functional value is
\(f(x)= 3x^2-2x+4=3.1^2-2.1+4=\) 5
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 1^-} f(x)= \lim_{x \to 1} 3x^2-2x+4=3.1^2-2.1+4=\) 5
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 1^+} f(x)= \lim_{x \to 1} 3x^2-2x+4=3.1^2-2.1+4=\) 5
Since,
LHL=RHL=f(x)
The function is continuous at x=1
This completes the solution
- \( f(x)=\frac{1}{2x}\) at x=0
Solution 👉 Click Here
Given that
\( f(x)=\frac{1}{2x} \)
At x=0, we compute the following
The functional value is
\(f(x)= \frac{1}{2x}=\frac{1}{2.(0)}=\) ∞
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 0^-} f(x)= \lim_{h \to 0^-} \frac{1}{2(0-h)}= \lim_{h \to 0} -\frac{1}{2h}=\) -∞
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 0^+} f(x)= \lim_{h \to 0^+} \frac{1}{2(0+h)}= \lim_{h \to 0} \frac{1}{2h}=\) ∞
Since,
LHL,RHL,f(x) are NOT equal
The function is NOT continuous at x=0
This completes the solution
- \( f(x)=\frac{1}{x-2}\) at x≠2
Solution 👉 Click Here
Given that
\( f(x)=\frac{1}{x-2} \)
At x=2, we compute the following
The functional value is
\(f(x)= \frac{1}{x-2}=\) NOT defined
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{h \to 0^-} \frac{1}{(2-h)-2}= \lim_{h \to 0} \frac{1}{-h}=\) -∞
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{h \to 0^+} \frac{1}{(2+h)-2}= \lim_{h \to 0} \frac{1}{h}=\) ∞
Since,
LHL ≠RHL,f(x) is NOT defined
The function is NOT continuous at x=2
This completes the solution
- \( f(x)=\frac{1}{3x}\) at x≠0
Solution 👉 Click Here
Given that
\( f(x)=\frac{1}{3x} \)
At x=0, we compute the following
The functional value is
\(f(x)= \frac{1}{3x}=\) NOT defined
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 0^-} f(x)= \lim_{h \to 0^-} \frac{1}{3(0-h)}= \lim_{h \to 0} -\frac{1}{3h}=\) -∞
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 0^+} f(x)= \lim_{h \to 0^+} \frac{1}{3(0+h)}= \lim_{h \to 0} \frac{1}{3h}=\) ∞
Since,
LHL ≠RHL,f(x) is NOT defined
The function is NOT continuous at x=0
This completes the solution
- \( f(x)=\frac{1}{1-x}\) at x=1
Solution 👉 Click Here
Given that
\( f(x)=\frac{1}{1-x} \)
At x=1, we compute the following
The functional value is
\(f(x)= \frac{1}{1-x}=\) ∞
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 1^-} f(x)= \lim_{h \to 0^-} \frac{1}{1-(1-h)}= \lim_{h \to 0} \frac{1}{h}=\) ∞
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 1^+} f(x)= \lim_{h \to 0^+} \frac{1}{1-(1+h)}= \lim_{h \to 0} \frac{1}{-h}=\) -∞
Since,
LHL ,RHL,f(x) are NOT equal
The function is NOT continuous at x=1
This completes the solution
- \( f(x)=\frac{1}{x-3}\) at x=3
Solution 👉 Click Here
Given that
\( f(x)=\frac{1}{x-3} \)
At x=3, we compute the following
The functional value is
\(f(x)= \frac{1}{x-3}=\) ∞
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 3^-} f(x)= \lim_{h \to 0^-} \frac{1}{(3-h)-3}= \lim_{h \to 0} \frac{1}{-h}=\) -∞
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 3^+} f(x)= \lim_{h \to 0^+} \frac{1}{(3+h)-3}= \lim_{h \to 0} \frac{1}{h}=\) ∞
Since,
LHL ≠RHL,f(x) are NOT equal
The function is NOT continuous at x=3
This completes the solution
- \( f(x)=\frac{x^2-9}{x-3}\) at x=3
Solution 👉 Click Here
Given that
\( f(x)=\frac{x^2-9}{x-3}\)
At x=3, we compute the following
The functional value is
\(f(x)= \frac{x^2-9}{x-3}=x+3=3+3=\) 6
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 3^-} f(x)= \lim_{x \to 3^-} \frac{x^2-9}{x-3}=\lim_{x \to 3}x+3=3+3=\) 6
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 3^+} f(x)= \lim_{x \to 3^+} \frac{x^2-9}{x-3}=\lim_{x \to 3}x+3=3+3=\) 6
Since,
LHL=RHL=f(x)
The function is continuous at x=3
This completes the solution
- \( f(x)=\frac{x^2-16}{x-4}\) at x=4
Solution 👉 Click Here
Given that
\( f(x)=\frac{x^2-16}{x-4}\)
At x=4, we compute the following
The functional value is
\(f(x)= \frac{x^2-16}{x-4}=x+4=4+4=\) 8
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 4^-} f(x)= \lim_{x \to 4^-} \frac{x^2-16}{x-4}=\lim_{x \to 4}x+4=4+4=\) 8
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 4^+} f(x)= \lim_{x \to 4^+} \frac{x^2-9}{x-3}=\lim_{x \to 4}x+4=4+4=\) 8
Since,
LHL=RHL=f(x)
The function is continuous at x=4
This completes the solution
- \( f(x)=\frac{|x-2|}{x-2}\) at x=2
Solution 👉 Click Here
Given that
\( f(x)=\frac{|x-2|}{x-2}\)
At x=2, we compute the following
The functional value is
\(f(x)=\frac{|x-2|}{x-2}=\) NOT defined
The left hand limit is
\(LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2} \frac{-(x-2)}{x-2} = \) -1
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2} \frac{(x-2)}{x-2}=\) 1
Since,
LHL≠RHL, f(x) is NOT defined
The function is NOT continuous at x=2
This completes the solution
- \( f(x)=\frac{x}{|x|}\) at x=0
Solution 👉 Click Here
Given that
\( f(x)=\frac{x}{|x|}\)
At x=0, we compute the following
The functional value is
\(f(x)=\frac{x}{|x|}=\) NOT defined
The left hand limit is
\(LHL=\displaystyle \lim_{x \to 0^-} f(x)= \lim_{x \to 0} \frac{x}{-x} = \) -1
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 0^+} f(x)= \lim_{x \to 0} \frac{x}{x}=\) 1
Since,
LHL≠RHL, f(x) is NOT defined
The function is NOT continuous at x=2
This completes the solution
- Discuss the continuity of functions at the points specified
- \( f(x)= \begin{cases} 2-x^2 & \text{for } x \le 2 \\ 1 & \text{for } x > 2 \end{cases} \) at x=2
Solution 👉 Click Here
Given that
\( f(x)=\begin{cases} 2-x^2 & \text{for } x \le 2 \\ 1 & \text{for } x > 2 \end{cases}\)
At x=2, we compute the following
The functional value is
\(f(x)=2-x^2=2-2^2=\) -2
The left hand limit is
\(LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2} 2-x^2=2-2^2=\) -2
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2} 1=\) 1
Since,
LHL,RHL, f(x) are NOT equal
The function is NOT continuous at x=2
This completes the solution
- \( f(x)= \begin{cases} 2x^2+4 & \text{for } x \le 2 \\ 4x+1 & \text{for } x > 2 \end{cases} \) at x=2
Solution 👉 Click Here
Given that
\( f(x)=\begin{cases} 2x^2+4 & \text{for } x \le 2 \\ 4x+1 & \text{for } x > 2 \end{cases}\)
At x=2, we compute the following
The functional value is
\(f(x)=2x^2+4=2.2^2+4=\) 12
The left hand limit is
\(LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2} 2x^2+4=2.2^2+4=\) 12
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2} 4x+1=4.2+1=\) 9
Since,
LHL,RHL, f(x) are NOT equal
The function is NOT continuous at x=2
This completes the solution
- \( f(x)= \begin{cases} 2x & \text{for } x \le 3 \\ 3x-3 & \text{for } x > 3 \end{cases} \) at x=3
Solution 👉 Click Here
Given that
\( f(x)=\begin{cases} 2x & \text{for } x \le 3 \\ 3x-3 & \text{for } x > 3 \end{cases}\)
At x=3, we compute the following
The functional value is
\(f(x)=2x=2.3=\) 6
The left hand limit is
\(LHL=\displaystyle \lim_{x \to 3^-} f(x)= \lim_{x \to 3} 2x=2.3=\) 6
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 3^+} f(x)= \lim_{x \to 3} 3x-3=3.3-3=\) 6
Since,
LHL,RHL, f(x) are equal
The function is continuous at x=3
This completes the solution
- \( f(x)= \begin{cases} 2x+1 & \text{for } x < 1 \\ 2 & \text{for } x =1 \\ 3x & \text{for } x >1 \end{cases} \) at x=1
Solution 👉 Click Here
Given that
\( f(x)=\begin{cases} 2x+1 & \text{for } x < 1 \\ 2 & \text{for } x =1 \\ 3x & \text{for } x >1 \end{cases}\)
At x=1, we compute the following
The functional value is
\(f(x)=\) 2
The left hand limit is
\(LHL=\displaystyle \lim_{x \to 1^-} f(x)= \lim_{x \to 1} 2x+1=2.1+1=\) 3
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 1^+} f(x)= \lim_{x \to 1} 3x=3.1=\) 3
Since,
LHL,RHL, f(x) are NOT equal
The function is NOT continuous at x=1
This completes the solution
- A function is defined as follows
\( f(x)= \begin{cases} x^2+2 & \text{for } x < 5 \\ 20 & \text{for } x =5 \\ 3x+12 & \text{for } x > 5 \end{cases} \).
Show that f(x) has removable discontinuity at x=5
Solution 👉 Click Here
Given that
\( f(x)=\begin{cases} x^2+2 & \text{for } x < 5 \\ 20 & \text{for } x =5 \\ 3x+12 & \text{for } x > 5 \end{cases} \)
At x=5, we compute the following
The functional value is
\(f(x)=\) 20
The left hand limit is
\(LHL=\displaystyle \lim_{x \to 5^-} f(x)= \lim_{x \to 5} x^2+2=5^2+2=\) 27
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 5^+} f(x)= \lim_{x \to 5} 3x+12=3.5+12=\) 27
Since,
LHL=RHL≠f(x)
The function has removable discontinuity at x=5
This completes the solution
- A function is defined as follows
\( f(x)= \begin{cases} 2x-3 & \text{for } x < 2 \\ 2 & \text{for } x =2 \\ 3x-5 & \text{for } x > 2 \end{cases} \).
Is the function f(x) continuous at x=2? If not, how can the function f(x) be made continuous at x=2?
Solution 👉 Click Here
Given that
\( f(x)=\begin{cases} 2x-3 & \text{for } x < 2 \\ 2 & \text{for } x =2 \\ 3x-5 & \text{for } x > 2 \end{cases} \)
At x=2, we compute the following
The functional value is
\(f(x)=\) 2
The left hand limit is
\(LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2} 2x-3=2.2-3=\) 1
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2} 3x-5=3.2-5=\) 1
Since,
LHL=RHL≠f(x)
The function has removable discontinuity at x=2, so the function f(x) be made continuous at x=2 be redefining the function as below.
\( f(x)= \begin{cases} 2x-3 & \text{for } x < 2 \\ 1 & \text{for } x =2 \\ 3x-5 & \text{for } x > 2 \end{cases} \).
This completes the solution
- A function is defined as follows
\( f(x)= \begin{cases} kx+3 & \text{for } x \ge 2 \\ 3x-1 & \text{for } x < 2 \end{cases} \).
Find the value of k so that f(x) is continuous at x=2
Solution 👉 Click Here
Given that
\( f(x)=\begin{cases} kx+3 & \text{for } x \ge 2 \\ 3x-1 & \text{for } x < 2 \end{cases} \)
At x=2, we compute the following
The functional value is
\(f(x)=kx+3=\) 2k+3
The left hand limit is
\(LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2} 3x-1=3.2-1=\) 5
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2} kx+3=\) 2k+3
Since the functinuous at x=2, we must have
LHL=RHL=f(x)
or 2k+3=5
or k=1
This completes the solution
- A function is defined as follows
\( f(x)= \begin{cases} \frac{2x^2-18}{x-3} & \text{for } x \ne 3 \\ k & \text{for } x =3 \end{cases} \).
Find the value of k so that f(x) is continuous at x=3
Solution 👉 Click Here
Given that
\( f(x)=\begin{cases} \frac{2x^2-18}{x-3} & \text{for } x \ne 3 \\ k & \text{for } x =3 \end{cases} \)
At x=3, we compute the following
The functional value is
\(f(x)=\) k
The left hand limit is
\(LHL=\displaystyle \lim_{x \to 3^-} f(x)= \lim_{x \to 3} \frac{2x^2-18}{x-3}=2{x+3}=2{3+3}=\) 12
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 3^+} f(x)= \lim_{x \to 3} \frac{2x^2-18}{x-3}=2{x+3}=2{3+3}=\) 12
Since the functinuous at x=3, we must have
LHL=RHL=f(x)
or k=12
This completes the solution
Additional Questions [BCB, page 394]
- Define the continuity of a function at a point. Give with reason, an example of a continuouss function at a point. Is the function \(f(x)=\frac{1}{1-x}\) continuous at the point x=1?
Solution 👉 Click Here
Definition
Let 𝑓(𝑥) be a function defined at all values in an open interval containing a
Then we say that f(x) is continuous at a if
for a given number \(\epsilon > 0\), there exists a number \(\delta > 0\)such that
|f(x)-f(a)|<\(\epsilon\) whenever |x-a|<\(\delta\)
OR
the function f(x) is continuous at a if the following three conditions holds
- f(a) is defined
- \( \displaystyle \lim_{x\to a^{-}}f(x)\) exist = LHL
- \( \displaystyle \lim_{x\to a^{+}}f(x) \) exist = RHL
- LHL = RHL = f(a)
Example with reason
Let \( f(x)= \begin{cases} -x^2 +2 &\text { for } x \ne 2 \\ -2 &\text { for } x=2 \end{cases} \) thn f(x) is continuous at x=2.
Because
At \( x=2^{-}\), the left hand limit is
\( \displaystyle \lim_{x \to 2^{-}} f(x)= \lim_{x \to 2^{-} } -x^2+2 =-2\)
LHL
At \( x=2^{+}\), the right hand limit is
\( \displaystyle \lim_{x \to 2^{+}} f(x)= \lim_{x \to 2^{+}} = -x^2+2 =-2\)
RHL
At x=2, the value of the function is
f(2) = -2
Functional Value
Since, limiting value and functional value are equal, f(x) is continuous at x=2.
- When a function f(x) is said to be continuous at a point x=a? Discuss the continuity of \(f(x)=\begin{cases} x^2+2 & \text {for } x \le 5 \\ 3x+12 & \text{for } x > 5 \end{cases}\) at x=5
Solution 👉 Click Here
Continuous
Let 𝑓(𝑥) be a function defined at all values in an open interval containing a
Then we say that f(x) is continuous at a if
for a given number \(\epsilon > 0\), there exists a number \(\delta > 0\)such that
|f(x)-f(a)|<\(\epsilon\) whenever |x-a|<\(\delta\)
OR
the function f(x) is continuous at a if the following three conditions holds
- f(a) is defined
- \( \displaystyle \lim_{x\to a^{-}}f(x)\) exist = LHL
- \( \displaystyle \lim_{x\to a^{+}}f(x) \) exist = RHL
- LHL = RHL = f(a)
Problem Solving
Given \(f(x)=\begin{cases} x^2+2 & \text {for } x \le 5 \\ 3x+12 & \text{for } x > 5 \end{cases}\) at x=5.
Here
At \( x=5^{-}\), the left hand limit is
\( \displaystyle \lim_{x \to 5^{-}} f(x)= \lim_{x \to 5^{-} } x^2+2 =27\)
LHL
At \( x=5^{+}\), the right hand limit is
\( \displaystyle \lim_{x \to 5^{+}} f(x)= \lim_{x \to 5^{+}} = 3x+12 =27\)
RHL
At x=5, the value of the function is
\(f(2) =x^2+2= 27\)
Functional Value
Since, limiting value and functional value are equal, f(x) is continuous at x=5.
- At what point is the function \(f(x)=\frac{x+1}{(x-2)(x-3)}\) (i) discontinuous (ii) continuous ?
Solution 👉 Click Here
Solution
The function \(f(x)=\frac{x+1}{(x-2)(x-3)}\) is (i) discontinuous at x=2 and x=3 (ii) is continuous in other points
- Doiscuss the continuity of the function f(x) at the point x=0.
\(f(x)=\begin{cases} x & \text {for } x > 0 \\ 1 & \text{for } x=0 \\ -x & \text{for } x < 0 \end{cases}\)
Solution 👉 Click Here
Given \(f(x)=\begin{cases} x & \text {for } x > 0 \\ 1 & \text{for } x=0 \\ -x & \text{for } x < 0 \end{cases}\)
Here
At \( x=0^{-}\), the left hand limit is
\( \displaystyle \lim_{x \to 0^{-}} f(x)= \lim_{x \to 0^{-} } -x =0\)
LHL
At \( x=0^{+}\), the right hand limit is
\( \displaystyle \lim_{x \to 0^{+}} f(x)= \lim_{x \to 0^{+}} = x =0\)
RHL
At x=0, the value of the function is
\(f(0) =1\)
Functional Value
Since, limiting value and functional value are NOT equal, f(x) is NOT continuous at x=0.
- A function f(x) is defind as follows. \(f(x)=\begin{cases} 2x+1 & \text {for } x < 1 \\ 2 & \text{for } x=1 \\ 3x & \text{for } x > 0 \end{cases}\)
Calculate the left hand limit and right hand limit of f(x) at x=1. Is the function continuous at x=1?
Solution 👉 Click Here
Given \(f(x)=\begin{cases} 2x+1 & \text {for } x < 1 \\ 2 & \text{for } x=1 \\ 3x & \text{for } x > 0 \end{cases}\)
Here
At \( x=1^{-}\), the left hand limit is
LHL=\( \displaystyle \lim_{x \to 1^{-}} f(x)= \lim_{x \to 1^{-} } 2x+1 =\)
3
At \( x=1^{+}\), the right hand limit is
RHL=\( \displaystyle \lim_{x \to 1^{+}} f(x)= \lim_{x \to 1^{+}} = 3x =\)
3
At x=1, the value of the function is
F=\(f(1) =\)
2
Since, limiting value and functional value are NOT equal, f(x) is NOT continuous at x=1.
- What do you understand by the limit of a function? Let a function f(x) is defined by
\(f(x)=\begin{cases} 2-x^2 & \text {for } x < 2 \\ 3 & \text{for } x=2 \\ x-4 & \text{for } x > 2 \end{cases}\)
Verify that the limit of the function f(x) exists at x=2. Is the function f(x) continuous at x=2? If not why? State how can you make it continuous.
Solution 👉 Click Here
Given \(f(x)=\begin{cases} 2-x^2 & \text {for } x < 2 \\ 3 & \text{for } x=2 \\ x-4 & \text{for } x > 2 \end{cases}\)
Here
At \( x=2^{-}\), the left hand limit is
LHL=\( \displaystyle \lim_{x \to 2^{-}} f(x)= \lim_{x \to 2^{-} } 2-x^2 =\)
-2
At \( x=1^{+}\), the right hand limit is
RHL=\( \displaystyle \lim_{x \to 2^{+}} f(x)= \lim_{x \to 2^{+}} = x-4=\)
-2
Since,LHL=RHL, the limit of the function f(x) exists at x=2
Next,
At x=2, the value of the function is
F=\(f(2) =\)
3
The function f(x) is NOT continuous at x=2. Because, limiting value and functional value are NOT equal.
We can make the function f(x) continuous by re-defining it as below.
\(f(x)=\begin{cases} 2-x^2 & \text {for } x < 2 \\ -2 & \text{for } x=2 \\ x-4 & \text{for } x > 2 \end{cases}\)
- A function f(x) is defined as under \(f(x)=\begin{cases} \frac{x^2-x-6}{x^2-2x-3} & \text {for } x \ne 3 \\ \frac{5}{3} & \text{for } x =3 \end{cases}\)
Prove that f(x) is discontinuous at x=3. Can the definition of f(x) for x=3 be modified so as to make it continuous there?
Solution 👉 Click Here
Given \(f(x)=\begin{cases} \frac{x^2-x-6}{x^2-2x-3} & \text {for } x \ne 3 \\ \frac{5}{3} & \text{for } x =3 \end{cases}\)
Here
At \( x=3^{-}\), the left hand limit is
LHL=\( \displaystyle \lim_{x \to 3^{-}} f(x)= \lim_{x \to 3^{-} } \frac{x^2-x-6}{x^2-2x-3}=\lim_{x \to 3^{-} }\frac{(x-3)(x+2)}{(x-3)(x+1)} =\lim_{x \to 3^{-} }\frac{x+2}{x+1}=\)
\(\frac{5}{4}\)
At \( x=3^{+}\), the right hand limit is
RHL=\( \displaystyle \lim_{x \to 3^{+}} f(x)= \lim_{x \to 3^{+}} \frac{x^2-x-6}{x^2-2x-3}=\lim_{x \to 3^{+} }\frac{(x-3)(x+2)}{(x-3)(x+1)} =\lim_{x \to 3^{+} }\frac{x+2}{x+1}=\)
\(\frac{5}{4}\)
At x=3, the value of the function is
F=\(f(3) =\)
\(\frac{5}{3}\)
The function f(x) is discontinuous at x=3. Because, limiting value and functional value are NOT equal.
We can make the function f(x) continuous by re-defining it as below.
\(f(x)=\begin{cases} \frac{x^2-x-6}{x^2-2x-3} & \text {for } x \ne 3 \\ \frac{5}{4} & \text{for } x =3 \end{cases}\)
- A function f(x) is defined as follows
\(f(x)=\begin{cases} \frac{1}{2}+x & \text {for } 0 < x <\frac{1}{2} \\ \frac{1}{2} & \text{for } x =\frac{1}{2} \\ \frac{3}{2}-x & \text{for }
\frac{1}{2} < x < 1 \end{cases}\)
Show that f(x) has removable discontinuity at \(x=\frac{1}{2}\)
Solution 👉 Click Here
Given \(f(x)=\begin{cases} \frac{1}{2}+x & \text {for } 0 < x <\frac{1}{2} \\ \frac{1}{2} & \text{for } x =\frac{1}{2} \\ \frac{3}{2}-x & \text{for }
\frac{1}{2} < x < 1 \end{cases}\)
Here
At \( x=\frac{1}{2}^{-}\), the left hand limit is
LHL=\( \displaystyle \lim_{x \to \frac{1}{2}^{-}} f(x)= \lim_{x \to \frac{1}{2}^{-} } \frac{1}{2}+x=\)
1
At \( x=\frac{1}{2}^{+}\), the right hand limit is
RHL=\( \displaystyle \lim_{x \to \frac{1}{2}^{+}} f(x)= \lim_{x \to \frac{1}{2}^{+}} \frac{3}{2}-x=\)
1
At \(x=\frac{1}{2}\), the value of the function is
F=\(f(\frac{1}{2}) =\)
\(\frac{1}{2}\)
The function f(x) is discontinuous at \(x=\frac{1}{2}\).
Since, LHL=RHL ≠f(x), the f(x) has removable discontinuity at \(x=\frac{1}{2}\)
- A function f(x) is defined in (0,3) as follows
\(f(x)=\begin{cases} x^2 & \text {for } 0 < x <1 \\ x & \text{for } 1 \le x < 2 \\ \frac{1}{4}x^3 & \text{for }
2 \le x < 3 \end{cases}\)
Show that f(x) is continuous at x=1 and x=2
Solution 👉 Click Here
Given \(f(x)=\begin{cases} x^2 & \text {for } 0 < x <1 \\ x & \text{for } 1 \le x < 2 \\ \frac{1}{4}x^3 & \text{for }
2 \le x < 3 \end{cases}\)
test at x=1
Here
At \( x=1^{-}\), the left hand limit is
LHL=\( \displaystyle \lim_{x \to 1^{-}} f(x)= \lim_{x \to 1^{-} } x^2=\)
1
At \( x=1^{+}\), the right hand limit is
RHL=\( \displaystyle \lim_{x \to 1^{+}} f(x)= \lim_{x \to 1^{+}} x=\)
1
At \(x=1\), the value of the function is
F=\(f(1) =x=\)
1
Since, LHL=RHL=f(1), the function is continuous at x=1
test at x=2
Here
At \( x=2^{-}\), the left hand limit is
LHL=\( \displaystyle \lim_{x \to 2^{-}} f(x)= \lim_{x \to 2^{-} } x=\)
2
At \( x=2^{+}\), the right hand limit is
RHL=\( \displaystyle \lim_{x \to 2^{+}} f(x)= \lim_{x \to 2^{+}} \frac{1}{4}x^3=\)
2
At \(x=2\), the value of the function is
F=\(f(2) =\frac{1}{4}x^3=\)
2
Since, LHL=RHL=f(2), the function is continuous at x=2
- A function f(x) is defined as follows
\(f(x)=\begin{cases} 3+2x & \text {for } -\frac{3}{2} \le x <0 \\ 3-2x & \text{for } 0 \le x < \frac{3}{2} \\ -3-2x & \text{for }
x \ge \frac{3}{2} \end{cases}\)
Show that f(x) is continuous at x=0 and discontinuous at \(x=\frac{3}{2}\)
Solution 👉 Click Here
Given \(f(x)=\begin{cases} 3+2x & \text {for } -\frac{3}{2} \le x <0 \\ 3-2x & \text{for } 0 \le x < \frac{3}{2} \\ -3-2x & \text{for }
x \ge \frac{3}{2} \end{cases}\)
test at x=0
Here
At \( x=0^{-}\), the left hand limit is
LHL=\( \displaystyle \lim_{x \to 0^{-}} f(x)= \lim_{x \to 0^{-} } 3+2x=\)
3
At \( x=0^{+}\), the right hand limit is
RHL=\( \displaystyle \lim_{x \to 0^{+}} f(x)= \lim_{x \to 0^{+}} 3-2x=\)
3
At \(x=0\), the value of the function is
F=\(f(0) =3-2x=\)
3
Since, LHL=RHL=f(0), the function is continuous at x=0
test at x=3/2
Here
At \( x=\frac{3}{2}^{-}\), the left hand limit is
LHL=\( \displaystyle \lim_{x \to \frac{3}{2}^{-}} f(x)= \lim_{x \to \frac{3}{2}^{-} } 3-2x=\)
0
At \( x=\frac{3}{2}^{+}\), the right hand limit is
RHL=\( \displaystyle \lim_{x \to \frac{3}{2}^{+}} f(x)= \lim_{x \to \frac{3}{2}^{+}} -3-2x=\)
-6
At \(x=\frac{3}{2}\), the value of the function is
F=\(f(\frac{3}{2}) =-3-2x=\)
-6
Since, LHL≠RHL=f(3/2), the function is NOT continuous at x=3/2
- A function f(x) is defined as follows
\(f(x)=\begin{cases} 1 & \text {for } x >0 \\ 0 & \text{for } x=0 \\ -1 & \text{for }
x < 01\end{cases}\)
Show that f(x) isdiscontinuous at x=0.
Solution 👉 Click Here
Given \(f(x)=\begin{cases} 1 & \text {for } x >0 \\ 0 & \text{for } x=0 \\ -1 & \text{for }
x < 0\end{cases}\)
Here
At \( x=0^{-}\), the left hand limit is
LHL=\( \displaystyle \lim_{x \to 0^{-}} f(x)= \lim_{x \to 0^{-} } -1=\)
-1
At \( x=0^{+}\), the right hand limit is
RHL=\( \displaystyle \lim_{x \to 0^{+}} f(x)= \lim_{x \to 0^{+}} 1=\)
1
At \(x=0\), the value of the function is
F=\(f(0) =0=\)
0
Since, LHL≠RHL≠f(0), the function is NOT continuous at x=0
- In the following, determine the value of the constant so that the given function is continuous at the point mentioned.
- \(f(x)=\begin{cases} kx^2 & \text {for } x \le 2 \\ 3 & \text{for } x > 2\end{cases}\) at x=2
Solution 👉 Click Here
Given that
\(f(x)=\begin{cases} kx^2 & \text {for } x \le 2 \\ 3 & \text{for } x > 2\end{cases}\)
At x=2, we compute the following
The functional value is
\(f(x)=kx^2=\) 4k
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2^-} kx^2=\) 4k
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2^+}3=\) 3
Since,
Function is assumed continuous at x=2, thus we must have
LHL=RHL=f(x)
or 4k=3
or \(k=\frac{3}{4}\)
- \(f(x)=\begin{cases}ax+5 & \text {for } x \le 2 \\ x-1 & \text{for } x > 2\end{cases}\) at x=2
Solution 👉 Click Here
Given that
\(f(x)=\begin{cases}ax+5 & \text {for } x \le 2 \\ x-1 & \text{for } x > 2\end{cases}\)
At x=2, we compute the following
The functional value is
\(f(x)=ax+5=\) 2a+5
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2^-} ax+5=\) 2a+5
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2^+}x-1=\) 1
Since,
Function is assumed continuous at x=2, thus we must have
LHL=RHL=f(x)
or 2a+5=1
or \(a=2\)
- \(f(x)=\begin{cases} 2px+3 & \text {for } x <1 \\ 1-px^2& \text{for } x \ge 1\end{cases}\) at x=1
Solution 👉 Click Here
Given that
\(f(x)=\begin{cases} 2px+3 & \text {for } x <1 \\ 1-px^2& \text{for } x \ge 1\end{cases}\)
At x=1, we compute the following
The functional value is
\(f(x)=1-px^2=\) 1-p
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 1^-} f(x)= \lim_{x \to 1^-} 2px+3=\) 2p+3
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 1^+} f(x)= \lim_{x \to 1^+}1-px^2=\) 1-p
Since,
Function is assumed continuous at x=1, thus we must have
LHL=RHL=f(x)
or 1-p=2p+3
or \(p=-\frac{2}{3}\)
- \(f(x)=\begin{cases} \frac{x^2-9}{x-3} & \text {for } x \ne 3 \\ k & \text{for } x =3 \end{cases}\) at x=3
Solution 👉 Click Here
Given that
\(f(x)=\begin{cases} \frac{x^2-9}{x-3} & \text {for } x \ne 3 \\ k & \text{for } x =3 \end{cases}\)
At x=3, we compute the following
The functional value is
\(f(x)=k=\) k
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 3^-} f(x)= \lim_{x \to 3^-} \frac{x^2-9}{x-3}=\lim_{x \to 3^-} (x+3)=\) 6
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 3^+} f(x)= \lim_{x \to 3^+}\frac{x^2-9}{x-3}=\lim_{x \to 3^-} (x+3)=\) 6
Since,
Function is assumed continuous at x=3, thus we must have
LHL=RHL=f(x)
or k=6
- Let \(f(x)=\begin{cases} 2x & \text {for } x <2 \\ 2 & \text{for } x=2 \\ x^2 & \text{for }
x >2 \end{cases}\). Show that f(x) has removable discontinuity at x=2.
Solution 👉 Click Here
Given that
\(f(x)=\begin{cases} 2x & \text {for } x <2 \\ 2 & \text{for } x=2 \\ x^2 & \text{for }
x >2 \end{cases}\)
At x=2, we compute the following
The functional value is
\(f(x)=2=\) 2
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2^-} 2x=\) 4
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2^+}x^2=\) 4
Since,
LHL=RHL≠f(x)
The function f(x) has removable discontinuity at x=2.
- What condition is necessary for a function f(x) to be continuous at the point x=a? In what condition will f(x) be discontinuous at x=a?
Solution 👉 Click Here
Let 𝑓(𝑥) be a function defined at all values in an open interval containing a
Then we say that f(x) is continuous at a if
for a given number \(\epsilon > 0\), there exists a number \(\delta > 0\)such that
|f(x)-f(a)|<\(\epsilon\) whenever |x-a|<\(\delta\)
OR
the function f(x) is continuous at x=a if the following three conditions holds
- f(a) is defined
- \( \displaystyle \lim_{x\to a^{-}}f(x)\) exist = LHL
- \( \displaystyle \lim_{x\to a^{+}}f(x) \) exist = RHL
- LHL = RHL = f(a)
If one of the following condition occured, then the function f(x) will be discontinuous at x=a
- LHL = RHL, BUT f(a) is NOT defined [Removable Discontinuity]
- f(a)=LHL = RHL≠f(a)[Removable Discontinuity]
- f(a)=LHL≠ RHL[Jump Discontinuity]
- LHL≠ RHL=f(a)[Jump Discontinuity]
- LHL= RHL=∞ or LHL= RHL=-∞[Infinite Discontinuity]
- LHL=∞, RHL=-∞ or LHL=-∞, RHL=∞[Infinite Discontinuity]
- LHL ,RHL ,f(a) are NOT equal [Discontinuity]
- A function f(x) is defined as \(f(x)=\begin{cases} 1 & \text {for } x \ne 0 \\ 2 & \text{for } x =0 \end{cases}\). Find \(\displaystyle \lim_{x \to 0} f(x)\) if exists. Is the function continuous at x=0?
Solution 👉 Click Here
Given that
\(f(x)=\begin{cases} 1 & \text {for } x \ne 0 \\ 2 & \text{for } x =0 \end{cases}\)
At x=0, we compute the following
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 0^-} f(x)= \lim_{x \to 0^-} 1=\) 1
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 0^+} f(x)= \lim_{x \to 0^+}1=\) 1
Since,
LHL=RHL
the limit \(\displaystyle \lim_{x \to 0} f(x)\) exists, and it is 1.
Again
The functional value at x=0 is
\(f(x)=2=\) 2
Since,
LHL=RHL≠f(a)
The function f(x) is NOT continuity at x=0.
- Find the point of discontinuous of the following functions
- \(f(x)=\frac{x+1}{x-1}\)
Solution 👉 Click Here
The point of discontinuous of the function \(f(x)=\frac{x+1}{x-1}\) is \(x-1=0\)
x-1=0
or x=1
- \(f(x)=\frac{3x-1}{x^3-5x^2+6x}\)
Solution 👉 Click Here
The point of discontinuous of the function \(f(x)=\frac{3x-1}{x^3-5x^2+6x}\) is \(x^3-5x^2+6x=0\)
\(x^3-5x^2+6x=0\)
or \(x=0\) or \(x^2-5x^+6=0\)
or \(x=0\) or \((x-2)(x-3)=0\)
or \(x=0,2,3\)
- Test the continuity of a function f(x)=x2+1 at x=0
Solution 👉 Click Here
Given that
\( f(x)=x^2+1 \)
At x=0, we compute the following
The functional value is
\(f(x)= x^2+1 = 0^2+1=\) 1
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 0^-} f(x)= \lim_{x \to 0^-} x^2+1 = 0^2+1=\) 1
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 0^+} f(x)= \lim_{x \to 0^+} x^2+1 = 0^2+1=\) 1
Since,
LHL=RHL=f(x)
The function is continuous at x=0
This completes the solution
- Test the continuity of a function
\(
f(x)=
\begin{cases}
\frac{|x|}{x} & x \ne 2\\
x & x =2
\end{cases}
\)
at x=2.
Solution 👉 Click Here
Given that
\(
f(x)=
\begin{cases}
\frac{|x|}{x} & x \ne 2\\
x & x =2
\end{cases}
\)
At x=2, we compute the following
The functional value is
\(
f(x)= x=
\) 2
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 2^-} \frac{|x|}{x}= \lim_{x \to 2^-} \frac{x}{x} = \) 1
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 2^+} \frac{|x|}{x}= \lim_{x \to 2^+} \frac{x}{x} = \) 1
Since,
LHL=RHL≠f(x)
The function is NOT continuous at x=2
This completes the solution
- Test the continuity of a function
\(
f(x)=
\begin{cases}
4x-3 & x < 2\\
(x-3)^2 & x ≥ 2
\end{cases}
\)
at x=2.
Solution 👉 Click Here
Given that
\(
f(x)=
\begin{cases}
4x-3 & x < 2\\
(x-3)^2 & x ≥ 2
\end{cases}
\)
At x=2, we compute the following
The functional value is
\(f(x)= (x-3)^2= (2-3)^2=\) 1
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2^-} 4x-3 = 4.2-3=\) 5
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2^+}(x-3)^2= (2-3)^2=\) 1
Since,
LHL≠RHL=f(x)
The function is NOT continuous at x=2
This completes the solution
- Test the continuity of a function
\(
f(x)=
\begin{cases}
\frac{x^2-4}{x-2} & x \ne 2\\
4 & x =2
\end{cases}
\)
at x=2.
Solution 👉 Click Here
Given that
\(
f(x)=
\begin{cases}
\frac{x^2-4}{x-2} & x \ne 2\\
4 & x =2
\end{cases}
\)
At x=2, we compute the following
The functional value is
\(f(x)= \) 4
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2^-} \frac{x^2-4}{x-2}=x+2=2+2=\) 4
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2^+}\frac{x^2-4}{x-2}=x+2=2+2=\) 4
Since,
LHL=RHL=f(x)
The function is continuous at x=2
This completes the solution
- Test the continuity of a function
\(f(x)=\begin{cases}
\frac{1}{x} & x < -1 \\
\frac{x-1}{2} & -1 \le x < 1 \\
\sqrt{x} & x \ge 1 \\
\end{cases}
\)
at x=-1.
Solution 👉 Click Here
Given that
\(f(x)=\begin{cases}
\frac{1}{x} & x < -1 \\
\frac{x-1}{2} & -1 \le x < 1 \\
\sqrt{x} & x \ge 1 \\
\end{cases}
\)
At x=-1, we compute the following
The functional value is
\(f(x)=\frac{x-1}{2}=\frac{-1-1}{2}=\) -1
The left hand limit is
\( LHL=\displaystyle \lim_{x \to -1^-} f(x)= \lim_{x \to -1^-} \frac{x-1}{2}=\frac{-1-1}{2}=\) -1
The right hand limit is
\(RHL= \displaystyle \lim_{x \to -1^+} f(x)= \lim_{x \to -1^+}\frac{x-1}{2}=\frac{-1-1}{2}=\) -1
Since,
LHL=RHL=f(x)
The function is continuous at x=-1
This completes the solution
- Test the continuity of a function
\( f(x)=\begin{cases} x-2 & x < 1 \\
\sqrt{x} & x \ge 1 \\ \end{cases} \) at x=-1 and at x=1.
Solution 👉 Click Here
Testing the continuity of the function at x=-1
Given that
\( f(x)=\begin{cases} x-2 & x < 1 \\
\sqrt{x} & x \ge 1 \\ \end{cases} \)
At x=-1, we compute the following
The functional value is
\(f(x)=x-2=-1-2=\) -3
The left hand limit is
\( LHL=\displaystyle \lim_{x \to -1^-} f(x)= \lim_{x \to -1^-} x-2=-1-2=\) -3
The right hand limit is
\(RHL= \displaystyle \lim_{x \to -1^+} f(x)= \lim_{x \to -1^+}x-2=-1-2=\) -3
Since,
LHL=RHL=f(x)
The function is continuous at x=-1
This completes the solution
Testing the continuity of the function at x=1
Given that
\( f(x)=\begin{cases} x-2 & x < 1 \\
\sqrt{x} & x \ge 1 \\ \end{cases} \)
At x=1, we compute the following
The functional value is
\(f(x)=\sqrt{x}=\sqrt{1}=\) 1
The left hand limit is
\( LHL=\displaystyle \lim_{x \to 1^-} f(x)= \lim_{x \to 1^-} x-2=1-2=\) -1
The right hand limit is
\(RHL= \displaystyle \lim_{x \to 1^+} f(x)= \lim_{x \to 1^+}\sqrt{x}=\sqrt{1}=\) 1
Since,
LHL≠RHL=f(x)
The function is NOT continuous at x=1
This completes the solution
- For what value of k would make the function continuous in each case? Explain how you found your value.
-
\(
f(x)=
\begin{cases}
\frac{x^2-9}{x+3} & x\ne -3 \\
k & x=-3
\end{cases}
\)
at x=-3
Solution 👉 Click Here
Given that
\(
f(x)=
\begin{cases}
\frac{x^2-9}{x+3} & x\ne -3 \\
k & x=-3
\end{cases}
\)
At x=-3, we compute the following
The functional value is
\(f(x)=k=\) k
The left hand limit is
\( LHL=\displaystyle \lim_{x \to -3^-} f(x)= \lim_{x \to -3^-} \frac{x^2-9}{x+3} =x-3=-3-3=\) -6
The right hand limit is
\(RHL= \displaystyle \lim_{x \to -3^+} f(x)= \lim_{x \to -3^+}\frac{x^2-9}{x+3} =x-3=-3-3=\) -6
Since,
Function is assumed continuous at x=-3, thus we must have
LHL=RHL=f(x)
or k=-6
This completes the solution
-
\(
f(x)=
\begin{cases}
\frac{\sin (5\pi x)-1}{2x-1} & x\ne \frac{1}{2} \\
k & x= \frac{1}{2}
\end{cases}
\)
at \( x=\frac{1}{2}\)
Solution 👉 Click Here
Given that
\(
f(x)=
\begin{cases}
\frac{\sin (5\pi x)-1}{2x-1} & x\ne \frac{1}{2} \\
k & x= \frac{1}{2}
\end{cases}
\)
At x=1/2, we compute the following
The functional value is
\(f(x)=k=\) k
The left hand limit is
| LHL | =\(\displaystyle \lim_{x \to \frac{1}{2}^-} f(x)\) |
| =\(\displaystyle \lim_{x \to \frac{1}{2}^-}\frac{\sin (5 \pi x)-1}{2x-1} \)
=\(\displaystyle \lim_{x \to \frac{1}{2}^-}\frac{\sin (\pi x)-1 }{2x-1} \)
=\(\displaystyle \lim_{x \to \frac{1}{2}^-}\frac{\sin (\pi x)-\sin (\frac{\pi}{2}) }{2x-1} \)
=\(\displaystyle \lim_{x \to \frac{1}{2}^-}\frac{2 \cos \left ( \frac{\pi x+\frac{\pi}{2}}{2}\right ). \sin \left ( \frac{\pi x+\frac{\pi}{2}}{2}\right ) }{2x-1} \)
=\(\displaystyle \lim_{x \to \frac{1}{2}^-}\frac{2 \cos \frac{\pi}{4} (2x+1). \sin \frac{\pi}{4} (2x-1)}{2x-1} \)
=\(\displaystyle \lim_{x \to \frac{1}{2}^-}\frac{2 \cos \frac{\pi}{4} (2x+1)}.\frac{\sin \frac{\pi}{4} (2x-1)}{2x-1} \)
=\(\displaystyle \lim_{x \to \frac{1}{2}^-} 2 \cos \left ( \frac{\pi}{4} (2x+1) \right ).\frac{\sin \frac{\pi}{4} (2x-1)}{\frac{\pi}{4}(2x-1)}. \frac{\pi}{4} \)
=\(\displaystyle \lim_{x \to \frac{1}{2}^-}2 \cos \left ( \frac{\pi}{4} (2x+1) \right ). \displaystyle \lim_{x \to \frac{1}{2}^-} \frac{\sin \frac{\pi}{4} (2x-1)}{\frac{\pi}{2}(2x-1)}. \frac{\pi}{2} \)
=\(\displaystyle \lim_{x \to \frac{1}{2}^-} 2 \cos \left ( \frac{\pi}{4} (2x+1) \right ). \frac{\pi}{2} \)
=\(2 \cos (\frac{\pi}{4} .2). \frac{\pi}{4} \)
=\(2 \cos (\frac{\pi}{2}). \frac{\pi}{4} \)
=\(2 .0. \frac{\pi}{4} \)
=0
|
Similarly, the right hand limit is
\(RHL= \displaystyle \lim_{x \to \frac{1}{2}^+} f(x)=\) 0
Since,
Function is assumed continuous at x=1/2, thus we must have
LHL=RHL=f(x)
or k=0
This completes the solution
-
\(
f(x)=
\begin{cases}
\frac{e^x-1}{x} & x\ne 0 \\
k & x=0
\end{cases}
\)
at x=0
Solution 👉 Click Here
Given that
\(
f(x)=
\begin{cases}
\frac{e^x-1}{x} & x\ne 0 \\
k & x=0
\end{cases}
\)
At x=0, we compute the following
The functional value is
\(f(x)=k=\) k
The left hand limit is
\(\displaystyle \lim_{x\to 0^-}\frac{e^x-1 }{x} \)
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
Similarly, the left hand limit is
\(\displaystyle \lim_{x\to 0^-}\frac{e^x-1 }{x} \)
or 1
Since,
Function is assumed continuous at x=0, thus we must have
LHL=RHL=f(x)
or k=1
This completes the solution.
- Evaluate \( \displaystyle \lim_{x \to 4} \frac{(x-3)}{(4-x)(x+3)}\)
Solution
Solution 👉 Click Here
Given that
\( \displaystyle \lim_{x \to 4} \frac{(x-3)}{(4-x)(x+3)}\)
At x=4, we compute the following
The functional value is
\(f(x)=\displaystyle \lim_{x \to 4} \frac{(x-3)}{(4-x)(x+3)}=\) Not Defined
The left hand limit is
| LHL | =\(\displaystyle \lim_{x \to 4^-} f(x)\) |
| =\(\displaystyle \lim_{x \to 4^-} \frac{(x-3)}{(4-x)(x+3)}\)
=\(\displaystyle \lim_{x \to 4} \frac{1}{4-x } \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\)
=\(\displaystyle \lim_{h \to 0} \frac{1}{4-(4-h)} \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\)
=\(\displaystyle \lim_{hx \to 0} \frac{1}{4-4+h} \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\)
=\(\displaystyle \lim_{hx \to 0} \frac{1}{h} \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\)
= \(\infty\)
|
The left hand limit is
| LHL | =\(\displaystyle \lim_{x \to 4^+} f(x)\) |
| =\(\displaystyle \lim_{x \to 4^+} \frac{(x-3)}{(4-x)(x+3)}\)
=\(\displaystyle \lim_{x \to 4^+} \frac{1}{4-x } \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\)
=\(\displaystyle \lim_{h \to 0} \frac{1}{4-(4+h)} \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\)
=\(\displaystyle \lim_{hx \to 0} \frac{1}{4-4-h} \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\)
=\(\displaystyle \lim_{hx \to 0} \frac{1}{-h} \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\)
= \(- \infty\)
|
Since,
LHL≠RHL, and f(x) is NOT defined
limit does NOT exist. Function is NOT continuous at x=4
This completes the solution
Method 2
If we substitute 𝑥=4 in the function \( \frac{(x-3)}{(4-x)(x+3)}\) then we get \(\frac{1}{0}\), so we do as follows.
|
(-∞,-3) |
(-3,3) |
(3,4) |
(4,∞) |
| (x-3) |
- |
- |
+ |
+ |
| (4-x) |
+ |
+ |
+ |
- |
| (x+3) |
- |
+ |
+ |
+ |
| Result |
|
|
+ |
- |
So, LHS=+∞, and RHS=-∞, therefor limit does NOT exist.
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