Answer
$\lim_{h\to0}\dfrac{(2+h)^{3}-8}{h}=12$
Work Step by Step
$\lim_{h\to0}\dfrac{(2+h)^{3}-8}{h}$
Try to evaluate the limit applying direct substitution:
$\lim_{h\to0}\dfrac{(2+h)^{3}-8}{h}=\dfrac{(2+0)^{3}-8}{0}=\dfrac{0}{0}$ Indeterminate form
The limit could not be evaluated using direct substitution. Factor the numerator of the function and simplify:
$\lim_{h\to0}\dfrac{(2+h)^{3}-8}{h}=...$
$...=\lim_{h\to0}\dfrac{[(2+h)-2][(2+h)^{2}+(2+h)(2)+2^{2}]}{h}=...$
$...=\lim_{h\to0}\dfrac{(h)[(2+h)^{2}+(2)(2+h)+4]}{h}=...$
$...=\lim_{h\to0}[(2+h)^{2}+(2)(2+h)+4]=...$
Try direct substitution again to evaluate the limit:
$...=\lim_{h\to0}[(2+h)^{2}+(2)(2+h)+4]=...$
$...=(2+0)^{2}+(2)(2+0)+4=4+4+4=12$
