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Solution:

Let $x=0+\mathrm{h}$, when $x$ is tends to $0^{+}$

Since $x$ tends to 0 , $h$ will also tend to 0 .

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

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

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

$=\lim _{h \rightarrow 0^{+}} \sin \frac{1}{0+h}$

$=\sin \frac{1}{0}$

$=\sin \infty$

$=\infty$

Let $x=0-h$, when $x$ is tends to $0^{-}$

Since $x$ tends to 0, h will also tend to 0 .

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

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

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

$=\lim _{h \rightarrow 0^{-}} \sin \frac{1}{0-h}$

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

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

$=-\sin \infty$

$=-\infty$

Since,

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

$\therefore \lim _{x \rightarrow 0} \sin \frac{1}{x}$ does not exist.