Answer
$\lim_{x\to0}\dfrac{(4+x)^{3}-64}{x}=48$
The graph is shown below:
Work Step by Step
$\lim_{x\to0}\dfrac{(4+x)^{3}-64}{x}$
Try to evaluate the limit applying direct substitution:
$\lim_{x\to0}\dfrac{(4+x)^{3}-64}{x}=\dfrac{(4+0)^{3}-64}{0}=\dfrac{0}{0}$ Indeterminate form.
The limit could not be evaluated using direct substitution. Factor the numerator and simplify:
$\lim_{x\to0}\dfrac{(4+x)^{3}-64}{x}=...$
$...=\lim_{x\to0}\dfrac{[(4+x)-4][(4+x)^{2}+4(4+x)+4^{2}]}{x}=...$
$...=\lim_{x\to0}\dfrac{(x)[(4+x)^{2}+4(4+x)+16]}{x}=...$
$...=\lim_{x\to0}[(4+x)^{2}+4(4+x)+16]=...$
Try to evaluate the limit using direct substitution again:
$...=\lim_{x\to0}[(4+x)^{2}+4(4+x)+16]=...$
$...=(4+0)^{2}+4(4+0)+16=16+16+16=48$
