Answer
$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=2$$
Work Step by Step
$$f(x)=x^2\hspace{1cm} x=1$$
So $f(x+h)=(x+h)^2=x^2+2xh+h^2$
Therefore, $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{(x^2+2xh+h^2)-(x^2)}{h}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{2xh+h^2}{h}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}(2x+h)$$
Substitute $x=1$ here: $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}(2+h)$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=2+0=2$$
