Find the value

Question:

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

Solution:

Left Hand Limit(L.H.L.):

$\lim _{x \rightarrow 0^{-}} f(x)$

$=\lim _{x \rightarrow 0^{-}} e^{\frac{-1}{(-x)}}$

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

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

$=e^{\infty}$

$\lim _{x \rightarrow 0^{+}} f(x)$

Right Hand Limit(R.H.L.):

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

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

$=e^{-\infty}$

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

[ Formula $\frac{1}{\infty}=0$, anything to the power infinity is also infinity. Thus $\frac{1}{e \infty}=\frac{1}{\infty}=0$ ]

$=0$

Since

$\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$

$\therefore \lim _{x \rightarrow 0} \mathrm{e}^{-1 / x}$ does not exist.

 

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