Question

${lim}_{x\to 2}\frac{x^5-32}{x^3-8}$

Answer 1

We have,

${lim}_{x\to 2}\frac{x^5-32}{x^3-8}$

This is the $\frac{0}{0}$ form.

So, apply L-hospital rule

${lim}_{x\to 2}\frac{5x^4-0}{3x^2-0}$

${lim}_{x\to 2}\frac{5x^4}{3x^2}$

${lim}_{x\to 2}\frac{5x^2}{3}$

$=\frac{5\times 2^2}{3}$

$=\frac{20}{3}$

Hence, this is the answer.