## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=2$$
$$f(x)=x^2\hspace{1cm} x=1$$ 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: $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}(2+h)$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=2+0=2$$