Question

Let $f\left(x\right)=x^2\left(\frac{e^{\frac{1}{x}}-e^{\frac{-1}{x}}}{e^{\frac{1}{x}}+e^{\frac{-1}{x}}}\right),x\ne 0$, 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. $\underset{x\to 0}{lim}f\left(x\right)=1$

Answer 1

The correct option is A $\underset{x\to 0+}{lim}f\left(x\right)$ doesn't exist

We have,

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

$L.H.L.$

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

$={lim}_{x\to 0-}{x}^2\left(\frac{\frac{e^{\frac{1}{x}}{e}^{\frac{1}{x}}-1}{e^{\frac{1}{x}}}}{\frac{e^{\frac{1}{x}}{e}^{\frac{1}{x}}+1}{e^{\frac{1}{x}}}}\right)$

$={lim}_{x\to 0-}{x}^2\left(\frac{e^{\frac{1}{x}}{e}^{\frac{1}{x}}-1}{e^{\frac{1}{x}}{e}^{\frac{1}{x}}+1}\right)\therefore {lim}_{x\to 0-}{e}^{\frac{1}{x}}=0$

then,

$=0^2\left(\frac{0-1}{0+1}\right)$

$=0\times -1$

$=0$

$R.H.L.$

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

$={lim}_{x\to 0+}{x}^2\left(\frac{\frac{e^{\frac{1}{x}}{e}^{\frac{1}{x}}-1}{e^{\frac{1}{x}}}}{\frac{e^{\frac{1}{x}}{e}^{\frac{1}{x}}+1}{e^{\frac{1}{x}}}}\right)$

$={lim}_{x\to 0+}{x}^2\left(\frac{e^{\frac{1}{x}}{e}^{\frac{1}{x}}-1}{e^{\frac{1}{x}}{e}^{\frac{1}{x}}+1}\right)\therefore {lim}_{x\to 0+}{e}^{\frac{1}{x}}=\infty$

$then,$

$=0^2\left(\frac{\infty -1}{\infty +1}\right)$

$=0\times \infty$

$=\infty$

Then,

$L.H.L\ne R.H.L$

Hence, Limit does not exist.