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 $f(x)=\dfrac{1}{x}$ and $f(1)=1$.
$\lim\limits_{h\to0^+}\dfrac{f(h+1)-f(1)}{h}=\lim\limits_{h\to0^+}\dfrac{\dfrac{1}{h+1}-1}{h}=\lim\limits_{h\to0^+}\dfrac{-h}{h(h+1)}=\lim\limits_{h\to0^+}\dfrac{-1}{h+1}=-1...(1)$
The left-hand derivative of $f(x)$ at $P(1,1)$ is given as :
we have $f(x)=x$ and $f(1)=1$. Therefore,
This implies that $\lim\limits_{h\to0^-}\dfrac{f(h+1)-f(1)}{h}=\lim\limits_{h\to0^-}\dfrac{h+1-1}{h}=1 ....(2)$
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 $P$.
