Answer
The detailed explanations are below.
Work Step by Step
- The right-hand derivative of $f(x)$ at $P(1,1)$:
As $x\to1^+$, $f(x)=1/x$ and $f(1)=1$. Therefore, $$\lim_{h\to1^+}\frac{f(h+1)-f(1)}{h}=\lim_{h\to1^+}\frac{\frac{1}{h+1}-1}{h}=\lim_{h\to1^+}\frac{1-(h+1)}{h(h+1)}$$ $$=\lim_{h\to1^+}\frac{-h}{h(h+1)}=\lim_{h\to1^+}\frac{-1}{h+1}=-\frac{1}{2}$$
- The left-hand derivative of $f(x)$ at $P(1,1)$:
As $x\to1^-$, $f(x)=x$ and $f(1)=1$. Therefore, $$\lim_{h\to1^-}\frac{f(h+1)-f(1)}{h}=\lim_{h\to1^-}\frac{h+1-1}{h}=\lim_{h\to1^-}1=1$$
Since the left-hand derivative differs with the right-hand one, we conclude that $f(x)$ is not differentiable at $P$.
