Answer
The proof is detailed below.
Work Step by Step
- The right-hand derivative of $f(x)$ at $P(0,0)$:
As $x\to0^+$, $f(x)=x$. Therefore, $$\lim_{h\to0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to0^+}\frac{h-0}{h}=\lim_{h\to0^+}1=1$$
- The left-hand derivative of $f(x)$ at $P(0,0)$:
As $x\to0^-$, $f(x)=x^2$. Therefore, $$\lim_{h\to0^-}\frac{f(h)-f(0)}{h}=\lim_{h\to0^-}\frac{h^2-0}{h}=\lim_{h\to0^-}h=0$$
Since the left-hand derivative differs with the right-hand one, we conclude that $f(x)$ is not differentiable at $P$.
