Evaluate

To evaluate:

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

Formula used: We have,

$\frac{x^{m}-y^{m}}{x-y}=m y^{m-1}$

As $x \rightarrow 4$, we have

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

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

$\lim _{x \rightarrow 2} \frac{x^{5}-32}{x^{3}-8}=\lim _{x \rightarrow 2} \frac{5(2)^{4}}{3(2)^{2}}$

$\lim _{x \rightarrow 2} \frac{x^{5}-32}{x^{3}-8}=\frac{20}{3}$

Thus, the value of $\lim _{x \rightarrow 2} \frac{x^{5}-32}{x^{3}-8}$ is $\frac{20}{3}$