Question

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

Answer 1

Given

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

Left hand limit.

L.H.L $=\underset{x\to 0^{-}}{lim}\frac{e^{\frac{1}{0-}}-1}{e^{\frac{1}{0-}}+1}=\frac{e^{-\infty }-1}{e^{-\infty }+1}=\frac{0-1}{0+1}=-1$

we know that $e^{\infty }=\infty$

$e^{-\infty }=\frac{1}{e^{\infty }}=\frac{1}{\infty }\Rightarrow 0$

Right hand limit

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

$=\frac{1-0}{1+0}=1$

So, here $L.H.L\ne R.H.L$

So, limit does not exist.