Question

Find whether the limit $\underset{x\to 0}{lim}\left(\frac{e^{1/x}-1}{e^{1/x}+1}\right)$ exits or not ?

Answer 1

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

$L.H.L=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}\frac{e^{\frac{-1}{h}}-1}{e^{\frac{-1}{h}}+1}=\underset{h\to 0}{lim}\frac{(\frac{1}{e^{\frac{1}{h}}}-1)}{(\frac{1}{e^{\frac{1}{h}}}+1)}=\frac{0-1}{0+1}=-1$

$($as $h\to 0\Rightarrow \frac{1}{h}\to \infty \Rightarrow e^{\frac{1}{h}}\to \infty \Rightarrow \frac{1}{e^{\frac{1}{h}}}\to 0)$

$R.H.L=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}\frac{e^{\frac{1}{h}}-1}{e^{\frac{1}{h}}+1}=\underset{h\to 0}{lim}\frac{(1-\frac{1}{e^{\frac{1}{h}}})}{(1+\frac{1}{e^{\frac{1}{h}}})}=+1$

Clearly ${lim}_{x\to 0^{-}}f\left(x\right)\ne {lim}_{x\to 0^{+}}f\left(x\right).$

Therefore ${lim}_{x\to 0}f\left(x\right)$ does not exist.