Question

$\underset{x\to 0}{lim}\frac{e^{1/x}-1}{e^{1/x}+1}$, is

• A. $-1$
• B. $1$
• C. $0$
• D. Does not exist

Answer 1

The correct option is D Does not exist

Let $f\left(x\right)=\frac{e^{1/x}-1}{e^{1/x}+1}$. Then,
(LHL of $f\left(x\right)$ at x $=$ 0)
$=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}\frac{e^{-1/h}-1}{e^{-1/h}+1}$
$=\underset{h\to 0}{lim}\left(\frac{\frac{1}{e^{1/h}}-1}{\frac{1}{e^{1/h}}+1}\right)=-1$ $[e^{1/h}\to \infty \Rightarrow \frac{1}{e^{1/h}}\to 0]$
and,
[RHL of $f\left(x\right)$ at x $=$ 0]
$=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0+h)=\frac{e^{1/h}-1}{e^{1/h}+1}$
$=\underset{h\to 0}{lim}\left(\frac{1-\frac{1}{e^{1/h}}}{1+e^{1/h}}\right)$ [Dividing Nr and Dr by e${}^{1/h}$]
$=\frac{1-0}{1+0}=1$
Clearly, $\underset{x\to 0^{-}}{lim}f\left(x\right)\ne \underset{x\to 0^{+}}{lim}f\left(x\right).$
Hence, $\underset{x\to 0}{lim}$ f(x) does not exist.