Evaluate _ x 0 f(x), where f(x) = l l | x | x , & x 0 0 , & x = 0… - Vidyadip

Question

Evaluate $\mathop {\lim }\limits_{x \to 0}$ f(x), where f(x) = $\left\{ \begin{array} { l l } { \frac { | x | } { x } , } & { x \neq 0 } \\ { 0 , } & { x = 0 } \end{array} \right.$

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Answer

We have, f(x) = $\left\{ {\begin{array}{*{20}{l}} {\frac{{|x|}}{x},}&{x \ne 0} \\ {0,}&{x = 0} \end{array}} \right.$
LHL = $\mathop {\lim }\limits_{x \to {0^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {0^ - }} \frac{{|x|}}{x} = \mathop {\lim }\limits_{h \to 0} \frac{{|0 - h|}}{{(0 - h)}}$ [putting x = 0 - h as x $\rightarrow$ 0, then h $\rightarrow$ 0]
= $\mathop {\lim }\limits_{h \to 0} \frac{{| - h|}}{{ - h}}$
= $\mathop {\lim }\limits_{h \to 0} \frac{{ h}}{{ - h}}$ [$\because$ |-x| = x]
= - 1
RHL = $\mathop {\lim }\limits_{x \to {0^ + }}$ f(x) = $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{|x|}}{x} = \mathop {\lim }\limits_{h \to 0} \frac{{|0 + h|}}{{(0 + h)}} = \mathop {\lim }\limits_{h \to 0} \frac{h}{h}$ = 1 [putting x = 0 + h as x $\rightarrow$ 0, then h $\rightarrow$ 0]
$\mathop {\lim }\limits_{h \to {0^ - }}$ f(x) $\ne \mathop {\lim }\limits_{x \to {0^ + }}$ f(x)
Limit does not exist at x = 0

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