Consider f(x) = begin{cases} frac{x^2}{|x|}, | vedclass

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(B) We have $f(x) = \begin{cases} \frac{x^2}{|x|}, & x 
e 0 \ 0, & x = 0 \end{cases} = \begin{cases} x, & x > 0 \ 0, & x = 0 \ -x, & x < 0 \end{ca

Vedclass · Accepted Answer

$f(x)$ is continuous everywhere

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(B) We have $f(x) = \begin{cases} \frac{x^2}{|x|}, & x 
e 0 \ 0, & x = 0 \end{cases} = \begin{cases} x, & x > 0 \ 0, & x = 0 \ -x, & x We have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0} (-x) = 0$ and $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0} (x) = 0$.Since $f(0) = 0$,we have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = 0$.Thus,$f(x)$ is continuous at $x = 0$.Since $f(x)$ is continuous for all $x 
e 0$ as well,$f(x)$ is continuous everywhere.Now,checking differentiability at $x = 0$:$Lf'(0) = \lim_{h 	o 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{-h - 0}{h} = -1$.$Rf'(0) = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{h - 0}{h} = 1$.Since $Lf'(0) 
e Rf'(0)$,$f(x)$ is not differentiable at $x = 0$.

Consider $f(x) = \begin{cases} \frac{x^2}{|x|}, & x \ne 0 \\ 0, & x = 0 \end{cases}$

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