Answer
$\displaystyle\lim_{x\rightarrow 1} \dfrac{x^5+x-2}{x^2+x-2}=2$
Work Step by Step
We have to estimate the limit:
$\displaystyle\lim_{x\rightarrow 1\pm} \dfrac{x^5+x-2}{x^2+x-2}$
Graph the function:
Therefore we get:
$\displaystyle\lim_{x\rightarrow 1^{-}} \dfrac{x^5+x-2}{x^2+x-2}=2$
$\displaystyle\lim_{x\rightarrow 1^{+}} \dfrac{x^5+x-2}{x^2+x-2}=2$
As the left hand limit and the right hand limit are equal, we have:
$\displaystyle\lim_{x\rightarrow 1} \dfrac{x^5+x-2}{x^2+x-2}=2$
There is a hole in the graph in $(1,2)$.
