Evaluate

To evaluate: $\lim _{x \rightarrow-2} \frac{x^{3}+8}{x+2}$

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 $x \rightarrow-2$, we have

$\lim _{x \rightarrow-2} \frac{x^{3}+8}{x+2}=\lim _{x \rightarrow-2} \frac{(x+2)\left(x^{2}-2 x+4\right)}{x+2}$

$\lim _{x \rightarrow-2} \frac{x^{3}+8}{x+2}=\lim _{x \rightarrow-2}\left(x^{2}-2 x+4\right)$

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

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

Thus, the value of $\lim _{x \rightarrow-2} \frac{x^{3}+8}{x+2}$ is 4