Question

Which of the following function(s) not defined at $x = 0$ has non removable discontinuity at the origin?
A) $f(x) = \dfrac{1}{{1 + {2^{\cot x}}}}$
B) $f(x) = \cos \left( {\dfrac{{\left| {\sin x} \right|}}{x}} \right)$
C) $f(x) = x\sin \dfrac{\pi }{x}$
D) $f(x) = \dfrac{1}{{\ln \left| x \right|}}$

Answer 1

In this question we will see if the RHL=LHL which should not be equal to the function of 0. For that we will check every option to see if it fulfills the criteria of LHL=RHL. That is how we will find the answer.

Complete step by step solution:
We will first see what the criteria for finding the discontinuous function are
Criteria for discontinuity is
LHL=RHL $\ne f(0)$
Now let’s check our first function which is Option A.
$f(x) = \dfrac{1}{{1 + {2^{\cot x}}}}$
Now we will check in the limits of the function
LHL (Left hand limit)
$\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{1}{{1 + {2^{\cot x}}}}$
Applying the limits which is $x \to {0^ - }$
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{1}{{1 + {2^{\cot {0^ - }}}}}$
Now, we know that
$\cot ( - 0) = - \infty$
Therefore, we will get
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{1}{{1 + {2^{ - \infty }}}}$
So, we can see that
$\cot ( - 0.00001) = - \infty$
Hence, we will get
$\Rightarrow \dfrac{1}{{1 + {2^{ - \infty }}}} = 1$
This the LHL
Now we will see RHL (Right hand limit)
$\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{1}{{1 + {2^{\cot {0^ + }}}}}$
We will apply the limit
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{1}{{1 + {2^\infty }}}$
We will get
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{1}{\infty }$
As we know that
$\Rightarrow \dfrac{1}{\infty } = 0$
$LHL \ne RHL$
Therefore, the criteria didn’t apply.
Now we will check Option B
$f(x) = \cos \left( {\dfrac{{\left| {\sin x} \right|}}{x}} \right)$
Now let’s check LHL
$\mathop {\lim }\limits_{x \to {0^ - }} \cos \left( {\dfrac{{\left| {\sin x} \right|}}{x}} \right)$
Now for limits $x \to {0^ - }$
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} \cos \left( {\dfrac{{ - \sin x}}{x}} \right)$
As we know that
$\dfrac{{\sin x}}{x} = 1$
Therefore, we will get
$\cos ( - 1) = \cos 1$
Now we will see its RHL
Which limits is $x \to {0^ + }$ $$$$
$\mathop {\lim }\limits_{x \to {0^ + }} \cos \left( {\dfrac{{\left| {\sin x} \right|}}{x}} \right)$
Now we will put the value
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} \cos \left( {\dfrac{{\sin x}}{x}} \right)$
As we know already
$\dfrac{{\sin x}}{x} = 1$
So, we get
$\Rightarrow \cos (1) = 1$
Hence, we get
LHL=RHL
When we put $f\left( 0 \right)$
$\cos \left( {\dfrac{{\left| {\sin 0} \right|}}{0}} \right) \to$ not defined
We can see that
$\dfrac{0}{0}$ is not defined
$f(0) =$ Not defined
So, We, can see that
$LHL = RHL \ne f(0)$ is all being prove that it is a discontinuous function

Hence the answer is B.

Note:
Discontinuous function is the function which creates a discontinuous graph, which don’t have flow of lines in a graph , these types of functions are known as discontinuous functions.
Continuous functions are the functions which create a continuous graph , which have flow of the lines in a graph. These types of functions are known as the continuous function.