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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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Evaluate $\lim\limits_{x \to 0} f(x)$ where $f(x)=\left\{\begin{array}{1 1}\large\frac{|x|}{x},&x\neq 0\\0,&x=0\end{array}\right.$

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The given function is
$f(x)=\left\{\begin{array}{1 1}\large\frac{|x|}{x},&x\neq 0\\0,&x=0\end{array}\right.$
$ \lim\limits_{x \to 0^-}f(x) = \lim\limits_{x \to 0^-} \bigg[ \large\frac{|x|}{x} \bigg]$
$ = \lim\limits_{x \to 0} = \bigg( \large\frac{-x}{x} \bigg) $$ \quad \quad [ when\: x \: is \: negative, |x| = -x]$
$ = \lim\limits_{x \to 0} (-1)$
$ = -1$
$ \lim\limits_{x \to 0^+}f(x) = \lim\limits_{x \to 0^+} \bigg[ \large\frac{|x|}{x} \bigg]$
$ = \lim\limits_{x \to 0} = \bigg( \large\frac{x}{x} \bigg) $$ \quad \quad [ when\: x \: is \: positive, |x| = x]$
$ = \lim\limits_{x \to 0} (1)$
$ = 1$
It is observed that $ \lim\limits_{ x \to 0^-} f(x) \neq \lim\limits_{ x \to 0^+} f(x) $
Hence, $\lim\limits_{ x \to 0} f(x) $ does not exist.
answered Apr 5, 2014 by thanvigandhi_1
 

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