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Question

Check continuity $$f(x)=\begin{cases} \left| x \right| +1,\quad \quad \quad x<0 \\ 0,\quad \quad \quad \quad \quad \quad x=0 \\ \left| x \right| +1,\quad \quad \quad x>0 \end{cases}$$


Solution

at $$x=0$$
R.H.L     $$\mathop {\lim }\limits_{x \to 0 + h} f\left( x \right)$$
$$ = \mathop {\lim }\limits_{h \to 0} \left| h \right| + 1$$
$$ = \mathop {\lim }\limits_{h \to 0} h + 1$$
$$ = 1$$
L.H.L       $$\mathop {\lim }\limits_{x \to 0 - h} f\left( x \right)$$
$$ = \mathop {\lim }\limits_{h \to 0} \left| { - h} \right| + 1$$
$$ = \mathop {\lim }\limits_{h \to 0} h + 1$$
$$ = 1$$
and  $$f\left( 0 \right) = 0$$
Since R.H.L $$=$$ L.H.L 
so, limit exist at $$x=0$$ 
since R.H.l $$=$$ L.H.L $$ \ne $$ $$f\left( 0 \right)$$
Therefore $$f\left( x \right)$$ is not continuous at $$x=0$$

Mathematics

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