Answer
$\displaystyle\lim_{x\rightarrow -3^{-}} \dfrac{x^2}{x^2-9}=\infty$
$\displaystyle\lim_{x\rightarrow -3^{+}} \dfrac{x^2}{x^2-9}=-\infty$
The limit does not exist
Work Step by Step
We have to estimate the limit:
$\displaystyle\lim_{x\rightarrow -3\pm} \dfrac{x^2}{x^2-9}$
Graph the function:
Therefore we get:
$\displaystyle\lim_{x\rightarrow -3^{-}} \dfrac{x^2}{x^2-9}=\infty$
$\displaystyle\lim_{x\rightarrow -3^{+}} \dfrac{x^2}{x^2-9}=-\infty$
As the left hand limit and the right hand limit are not equal, the limit of the function in -3 does not exist. The function tends to $\infty$ when $x\rightarrow -3^{-}$ and to $-\infty$ when $x\rightarrow -3^{+}$. There is a vertical asymptote $x=-3$.
