Answer
128
Work Step by Step
Numerator at $x=8:\,\,\,x^{3}-64x=(8)^{3}-64(8)=0$
Denominator at $x=8:\,\,\,x-8=8-8=0$
The function has the indeterminate form $\frac{0}{0}$ at $x=7$.
Transforming algebraically and canceling the common factor, we have
$\frac{x^{3}-64x}{x-8}=\frac{(x-8)(x^{2}+8x)}{x-8}=x^{2}+8x$
Therefore,
$\lim\limits_{x \to 8}\frac{x^{3}-64x}{x-8}=\lim\limits_{x \to 8}x^{2}+8x=(8)^{2}+8(8)=128$
