Question

Let $f\left(x\right)=x^2\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}},x\ne 0$ and $f\left(0\right)=1$ then

• A. $\underset{x\to 0^{+}}{lim}f\left(x\right)$ doesn't exist
• B. $\underset{x\to 0}{lim}f\left(x\right)$ doesn't exist
• C. $\underset{x\to 0}{lim}f\left(x\right)$ exist
• D. $f$ is a continuous function

Answer 1

The correct option is C $\underset{x\to 0}{lim}f\left(x\right)$ exist

$f\left(x\right)=x^2\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}},x\ne 0$ ...... given

by dividing $e^{1/x}$ that can be written as

$f\left(x\right)=x^2\left(\frac{e^{2/x}-1}{e^{2/x}+1}\right)=x^2(1-\frac{2}{e^{2/x}+1})$

$=x^2-\frac{2x^2}{e^{2/x}+1}$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=0-\frac{2\times 0}{\infty }$ $as(x\to 0^{+};\frac{1}{x}\to +\infty )$

$=0$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=0-\frac{2\times 0}{\frac{1}{\infty }+1}=0$

LHL$=$RHL $\Rightarrow \underset{x\to 0}{lim}f\left(x\right)=0$, Limit exists

But given $f\left(0\right)=1\Rightarrow f\left(0\right)\ne \underset{x\to 0}{lim}f\left(x\right)$

$\therefore$ Discontinuous.