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

Hint:
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.