Answer
$\displaystyle\lim_{x\rightarrow -2^-} \dfrac{4x^2+7}{x^3+8}=-\infty$
$\displaystyle\lim_{x\rightarrow -2^+} \dfrac{4x^2+7}{x^3+8}=\infty$
The limit does not exist
Work Step by Step
We have to estimate the limit:
$\displaystyle\lim_{x\rightarrow -2\pm} \dfrac{4x^2+7}{x^3+8}$
Graph the function:
Therefore we get:
$\displaystyle\lim_{x\rightarrow -2^-} \dfrac{4x^2+7}{x^3+8}=-\infty$
$\displaystyle\lim_{x\rightarrow -2^+} \dfrac{4x^2+7}{x^3+8}=\infty$
As the left hand limit and the right hand limit are not equal, the limit of the function in -2 does not exist. The function tends to $-\infty$ when $x\rightarrow -2^-$ and to $\infty$ when $x\rightarrow -2^+$. There is a vertical asymptote $x=-2$.
