Calculus (3rd Edition)

$\displaystyle\lim_{x\rightarrow 1} \dfrac{x^5+x-2}{x^2+x-2}=2$
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 $(1,2)$.