Question

Show that $\underset{x\to 0}{lim}\left(\frac{e^{1/x}-1}{e^{1/x}+1}\right)$ does not exist.

Answer 1

$\underset{x\to 0}{lim}\frac{(e^{\frac{1}{x}}-1)}{(e^{\frac{1}{x}}+1)}$

$L.HL=\underset{x\to 0^{-}}{lim}f\left(x\right)$

$=\underset{n\to 0}{lim}f(0-n)$

$=\underset{n\to 0}{lim}\frac{e^{-\frac{1}{n}}-1}{e^{-\frac{1}{n}}+1}$

$=\underset{n\to 0}{lim}\frac{\frac{1}{e^{\frac{1}{n}}}-1}{\frac{1}{e^{\frac{1}{n}}}+1}=-1$

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

$=\underset{h\to 0}{lim}f(0+h)$

$\underset{h\to 0}{lim}\frac{e^{\frac{1}{n}}-1}{e^{\frac{1}{n}}+1}$

$=\underset{h\to 0}{lim}\frac{1-\frac{1}{e^{\frac{1}{n}}}}{1+\frac{1}{e^{\frac{1}{n}}}}$

$=1$

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

Hence, $\underset{x\to 0}{lim}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}$ Doesn't exist.