Answer
$\displaystyle\lim_{x\rightarrow 4^-} \dfrac{x+1}{x-4}=-\infty$
$\displaystyle\lim_{x\rightarrow 4^+} \dfrac{x+1}{x-4}=\infty$
The limit does not exist
Work Step by Step
We have to estimate the limit:
$\displaystyle\lim_{x\rightarrow \pm4} \dfrac{x+1}{x-4}$
Graph the function:
Therefore we get:
$\displaystyle\lim_{x\rightarrow 4^-} \dfrac{x+1}{x-4}=-\infty$
$\displaystyle\lim_{x\rightarrow 4^+} \dfrac{x+1}{x-4}=\infty$
As the left hand limit and the right hand limit are not equal, the limit of the function in 4 does not exist. The function tends to $-\infty$ when $x\rightarrow 4-$ and to $\infty$ when $x\rightarrow 4+$. There is a vertical asymptote $x=4$.
