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Which one of the following is correct in respect of the
function $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}}{|\mathrm{x}|}$ for $\mathrm{x} \neq 0$ and $\mathrm{f}(0)=0 ?$
Options:
function $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}}{|\mathrm{x}|}$ for $\mathrm{x} \neq 0$ and $\mathrm{f}(0)=0 ?$
Solution:
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Verified Answer
The correct answer is:
$\mathrm{f}(\mathrm{x})$ is continuous every where
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\frac{\mathrm{x}^{2}}{\mathrm{x}}, & \mathrm{x} \neq 0 \\ 0 & \mathrm{x}=0\end{array}\right.$
$=\left\{\begin{array}{ll}\frac{x^{2}}{x}=x, & x>0 \\ 0, & x=0 \\ \frac{x^{2}}{-x}=-x, & x < 0\end{array}\right.$
Now, $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}(-x)=0$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(x)=0$
and $f(0)=0$ So, $f(x)$ is continuous at $x=0$ Also, $\mathrm{f}(\mathrm{x})$ is continuous for all other values of $\mathrm{x}$. Hence, $\mathrm{f}(\mathrm{x})$ is continuous everywhere.
$=\left\{\begin{array}{ll}\frac{x^{2}}{x}=x, & x>0 \\ 0, & x=0 \\ \frac{x^{2}}{-x}=-x, & x < 0\end{array}\right.$
Now, $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}(-x)=0$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(x)=0$
and $f(0)=0$ So, $f(x)$ is continuous at $x=0$ Also, $\mathrm{f}(\mathrm{x})$ is continuous for all other values of $\mathrm{x}$. Hence, $\mathrm{f}(\mathrm{x})$ is continuous everywhere.
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