Question

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

Answer 1

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

Now,

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

Put $x=0+h\Rightarrow h=x$

as $x\to 0^{+}\Rightarrow x>0$ slightly

$\Rightarrow h>0\Rightarrow h>0^{+}$

$={lim}_{h\to 0^{+}}\frac{1}{e^{\frac{1}{0+h}}}=\frac{1}{e^{\frac{1}{0}}}=\frac{1}{\infty }=0$

And,

${lim}_{x\to 0^{-}}{e}^{\frac{-1}{e^x}}={lim}_{x\to 0^{-}}\frac{1}{e^{\frac{1}{x}}}$

Put $x=0-h\Rightarrow h=-x$

as $x\to 0^{-}\Rightarrow x<0$ slightly

$\Rightarrow -x>0\Rightarrow h>0$

$\Rightarrow h\to 0^{+}$

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

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

$\Rightarrow {lim}_{x\to 0^{+}}{e}^{\frac{-1}{x}}\ne {lim}_{x\to 0^{-}}{e}^{\frac{-1}{x}}$

$\therefore {lim}_{x\to 0^{-}}{e}^{\frac{-1}{x}}$ does not exist.