Question

The function  \[f\left( x \right) = \begin{cases}\frac{e^{1/x} - 1}{e^{1/x} + 1}, & x \neq 0 \\ 0 , & x = 0\end{cases}\]

 

विकल्प

• is continuous at x = 0

• is not continuous at x = 0

• is not continuous at x = 0, but can be made continuous at x = 0

• none of these

Answer 1

is not continuous at x = 0

Given:  

\[f\left( x \right) = \begin{cases}\frac{e^{1/x} - 1}{e^{1/x} + 1}, & x \neq 0 \\ 0 , & x = 0\end{cases}\]

We have

\[\lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \left( \frac{e^\frac{1}{x} - 1}{e^\frac{1}{x} + 1} \right)\]

If \[e^\frac{1}{x} = t\] , then

 \[x \to 0, t \to \infty\]

Also,

\[f\left( 0 \right) = 0\]

\[\therefore \lim_{x \to 0} f\left( x \right) \neq f\left( 0 \right)\]

Hence,

\[f\left( x \right)\] is discontinuous at  \[x = 0\] .