Answer
$2$
Work Step by Step
Consider $f(x)=x^2;x=1$
and $f(x+h)=(x+h)^2=x^2+2xh+h^2$
Now, we have
$\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to0}\dfrac{(x^2+2xh+h^2)-(x^2)}{h}=\lim\limits_{h\to0}\dfrac{2xh+h^2}{h}
$
Thus, $\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to0}(2x+h)$
Plug $x=1$ , we get
$\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to0}(2+h)=2$
