Evaluate

To evaluate:

$\lim _{x \rightarrow 4} \frac{x^{3}-64}{x^{2}-16}$

Formula used: We have,

$\lim _{x \rightarrow a} f(x)=f(a)$

and

$x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$

As $x \rightarrow 4$, we have

$\lim _{x \rightarrow 4} \frac{x^{3}-64}{x^{2}-16}=\lim _{x \rightarrow 4} \frac{(x-4)\left(x^{2}+4 x+16\right)}{(x+4)(x-4)}$

$\lim _{x \rightarrow 4} \frac{x^{3}-64}{x^{2}-16}=\lim _{x \rightarrow 4} \frac{\left(x^{2}+4 x+16\right)}{(x+4)}$

$\lim _{x \rightarrow 4} \frac{x^{3}-64}{x^{2}-16}=\frac{48}{8}$

$\lim _{x \rightarrow 4} \frac{x^{3}-64}{x^{2}-16}=6$

Thus, the value of

$\lim _{x \rightarrow 4} \frac{x^{3}-64}{x^{2}-16}$ is 6.