Question

Given that $f\left(x\right)=\frac{xg\left(x\right)}{\left|x\right|},g\left(0\right)=g^{\prime }\left(0\right)=0$ and $f$ is continuous at $x=0$, the value of $f^{\prime }\left(0\right)$ is

Answer 1

$f\left(x\right)=\frac{xg\left(x\right)}{|x|}\Rightarrow f\left(x\right)=\{\begin{array}{c}g\left(x\right)x\ge 0\\ -g\left(x\right)x<0\end{array}$

$f\left(x\right)=\underset{h\to 0}{lim}\frac{g(x+h)-g\left(x\right)}{x+h-x}\Rightarrow f^{\prime }\left(0\right)=\underset{h\to 0^{+}}{lim}\frac{g(0+h)-g\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{g\left(h\right)}{h}=g^{\prime }\left(0\right)=0$

Hence, correcr answer is $0$.