Answer
$f(x)$ is not differentiable at $P$
Work Step by Step
The right-hand derivative of $f(x)$ at $P(1,1)$ is given as:
we have $f(x)=2x-1$ and $f(1)=1$
$\lim\limits_{h\to0^+}\dfrac{f(h+1)-f(1)}{h}=\lim\limits_{h\to0^+}\dfrac{2(h+1)-1-1}{h}=\lim\limits_{h\to0^+}\dfrac{2h}{h}=2 ...(1)$
The right-hand derivative of $f(x)$ at $P(1,1)$ is given as:
we have $f(x)=\sqrt x$ and $f(1)=1$
$\lim\limits_{h\to0^-}\dfrac{f(h+1)-f(1)}{h}=\lim\limits_{h\to0^-}\dfrac{\sqrt{h+1}-1}{h}=\lim\limits_{h\to0^-}\dfrac{(\sqrt{h+1}-1)(\sqrt{h+1}+1)}{h(\sqrt{h+1}+1)}=\lim\limits_{h\to0^-}\dfrac{1}{\sqrt{h+1}+1}=\dfrac{1}{2}...(2)$
From the above equations (1) and (2), we conclude that the left-hand derivative is not equal to the right-hand derivative. This means that $f(x)$ is not differentiable at $P$.
