Answer
$\lim_{h\to0}\dfrac{(2+h)^{2}-4}{h}=4$
Work Step by Step
$\lim_{h\to0}\dfrac{(2+h)^{2}-4}{h}$
Try to evaluate the limit using direct substitution:
$\lim_{h\to0}\dfrac{(2+h)^{2}-4}{h}=\dfrac{(2+0)^{2}-4}{0}=\dfrac{0}{0}$ Indeterminate form
The limit could not be evaluated using direct substitution. Factor the numerator of the function and simplify:
$\lim_{h\to0}\dfrac{(2+h)^{2}-4}{h}=\lim_{h\to0}\dfrac{[(2+h)-2][(2+h)+2]}{h}=...$
$...=\lim_{h\to0}\dfrac{(h)(h+4)}{h}=\lim_{h\to0}(h+4)=...$
Try to evaluate the limit using direct substitution again:
$\lim_{h\to0}(h+4)=0+4=4$