Evaluate
$\lim _{x \rightarrow 2}\left(\frac{x^{5}-32}{x^{3}-8}\right)$
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 $\frac{20}{3}$
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