Science:Math Exam Resources/Courses/MATH104/December 2015/Question 10 (a)/Solution 1

We observe that $\lim \nolimits _{x\to 0^{-}}-{\tfrac {1}{x}}=\infty$ and that $\lim \nolimits _{x\to 0^{+}}-{\tfrac {1}{x}}=-\infty .$ Hence

${\begin{aligned}\lim _{x\to 0^{-}}f(x)&=\lim _{x\to 0^{-}}x^{2}+x-1+\lim _{x\to 0^{-}}-{\frac {1}{x}}=-1+\infty ={\color {blue}\infty },\\\lim _{x\to 0^{+}}f(x)&=\lim _{x\to 0^{+}}x^{2}+x-1+\lim _{x\to 0^{+}}-{\frac {1}{x}}=-1+-\infty ={\color {blue}-\infty }.\end{aligned}}$