Question

The function $f\left(x\right)=\{\begin{array}{c}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}\\ 0,x=0\end{array},x\ne 0$

• A. is continuous at $x=0$
• B. is not continuous at $x=0$
• C. is not continuous at $x=0$, but can be made continuous at $x=0$
• D. None of these

Answer 1

Answer: (B)

(is not continuous at $x=0$)

Given $f\left(x\right)=\{\begin{array}{c}\frac{e^{1/x}-1}{e^{1/x}+1}\\ 0,x=0\end{array},x\ne 0$
For a function f(x) to be continuous at $x=0$ if and only if
$LHL=RHL=f\left(0\right)$.
Now lets find: $RHL={lim}_{x\to 0^{+}}f\left(x\right)$
$={lim}_{h\to 0}f(0+h)$
$={lim}_{h\to 0}f\left(h\right)$
$={lim}_{h\to 0}\frac{e^{\frac{1}{h}}-1}{e^{\frac{1}{h}}+1}$ [Replace $x$ by $h$]
$={lim}_{h\to 0}\frac{1-e^{-\frac{1}{h}}}{1+e^{-\frac{1}{h}}}$
Since, as $h\to 0$ then $\frac{1}{h}\to \infty$
So, above expression can be written as
$=\frac{1-e^{-\infty }}{1+e^{-\infty }}$
$=\frac{1-0}{1+0}$ [Since, $e^{-\infty }=0$]
$=1$
Similar way,
we can evaluate, $LHL={lim}_{x\to 0^{-}}f\left(x\right)$
$={lim}_{h\to 0}f(0-h)$
$={lim}_{h\to 0}f(-h)$
$={lim}_{h\to 0}\frac{e^{-\frac{1}{h}}-1}{e^{-\frac{1}{h}}+1}$
$=\frac{e^{-\infty }-1}{e^{-\infty }+1}$
$=\frac{0-1}{0+1}$ [Since, $e^{-\infty }=0$]
$=-1$
Thus, L.H.L$\ne$ R.H.S
Therefore, $f\left(x\right)$ is not continuous at $x=0$