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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
Question
Find all points of discontinuity of $\mathrm{f}$, where $\mathrm{f}$ is defined by: $f(x)=\left\{\begin{array}{l}\frac{|x|}{x}, \text { if } x \neq 0 \\ 0, \text { if } x=0\end{array}\right.$
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Answer

Given: $f\left( x \right) = \left\{ \begin{gathered} \frac{{\left| x \right|}}{x},\,\,if\,x \ne 0 \hfill \\ 0,\,if\,x = 0 \hfill \\ \end{gathered} \right.$
i.e., $\frac{x}{x} = 1$ if $x > 0$ and $\frac{{ - x}}{x} = - 1$ if $x < 0.$
$\Rightarrow f\left( x \right) = 1$ if $x > 0, f(x) = - 1$ if $x < 0$ and $f(x) = 0 if x = $0
It is clear that domain of $f(x)$ is $R$ as $f(x)$ is defined for $x > 0, x < 0$ and $x = 0$.
For all $x > 0,f\left( x \right) = 1$ is a constant function and hence continuous.
For all $x < 0,f\left( x \right) = - 1$ is a constant function and hence continuous.
Therefore $f(x)$ is continuous on $R – {0}$.
Now Left Hand limit at $ x=0 = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - 1} \right) = - 1$
Right Hand limit at $ x=0 = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( 1 \right) = 1$
Since $\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)$
Therefore, $\mathop {\lim }\limits_{x \to 0} f\left( x \right)$ does not exist and hence $f(x)$ is discontinuous at $x = 0 \ ($only$)$.

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