Answer
The detailed explanations are below.
Work Step by Step
- The right-hand derivative of $f(x)$ at $P(1,1)$:
As $x\to1^+$, $f(x)=2x-1$ and $f(1)=1$. Therefore, $$\lim_{h\to1^+}\frac{f(h+1)-f(1)}{h}=\lim_{h\to1^+}\frac{2(h+1)-1-1}{h}=\lim_{h\to1^+}\frac{2h+2-2}{h}$$ $$=\lim_{h\to1^+}\frac{2h}{h}=\lim_{h\to1^+}2=2$$
- The left-hand derivative of $f(x)$ at $P(1,1)$:
As $x\to1^-$, $f(x)=\sqrt x$ and $f(1)=1$. Therefore, $$\lim_{h\to1^-}\frac{f(h+1)-f(1)}{h}=\lim_{h\to1^-}\frac{\sqrt{h+1}-1}{h}$$
Multiply both numerator and denominator by $\sqrt{h+1}+1$:
$$\lim_{h\to1^-}\frac{f(h+1)-f(1)}{h}=\lim_{h\to1^-}\frac{(\sqrt{h+1}-1)(\sqrt{h+1}+1)}{h(\sqrt{h+1}+1)}$$ $$=\lim_{h\to1^-}\frac{h+1-1}{h(\sqrt{h+1}+1)}=\lim_{h\to1^-}\frac{1}{\sqrt{h+1}+1}=\frac{1}{\sqrt{1+1}+1}=\frac{1}{1+\sqrt2}$$
Since the left-hand derivative differs with the right-hand one, we conclude that $f(x)$ is not differentiable at $P$.
