Let $f(x)=\left\{\begin{array}{l}\frac{|x-3|}{(x-3)^{\prime}} x \neq 3 \\ 0, \quad x=3\end{array}\right.$
Show that $\lim _{x \rightarrow 3} f(x)$ does not exist.
Left Hand Limit(L.H.L.):
$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}} \frac{|x-3|}{x-3}$
$=\lim _{x \rightarrow 3^{-}} \frac{-(x-3)}{x-3}$
$=\lim _{x \rightarrow 3^{-}}-1$
$=-1$
Right Hand Limit(R.H.L.):
$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}} \frac{|x-3|}{x-3}$
$=\lim _{x \rightarrow 3^{+}} \frac{(x-3)}{x-3}$
$=\lim _{x \rightarrow 3^{+}} 1$
$=1$
$\lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)$
Thus, $\lim _{x \rightarrow 3} f(x)$ does not exist.
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