Science:Math Exam Resources/Courses/MATH110/April 2019/Question 02 (a)(ii)/Solution 1

The denominator is zero for $x=-4,0,4$. In part (i) we saw that $x=4$ is not a vertical asymptote. The other two candidates are asymptotes.

$\lim _{x\to -4^{-}}{\frac {x-4}{x(x^{2}-16)}}=\infty ,$ since $\lim _{x\to -4^{-}}(x-4)=-8\neq 0$ and $\lim _{x\to -4^{-}}x(x^{2}-16)=0$.

$\lim _{x\to 0^{-}}{\frac {x-4}{x(x^{2}-16)}}=-\infty ,$ since $\lim _{x\to 0^{-}}x-4=-4\neq 0$ and $\lim _{x\to 0^{-}}x(x^{2}-16)=0$.

Since we have shown that function approaches infinity (or negative infinity, respectively) as $x\to -4$ and $x\rightarrow 0$ from the left, this shows that the function has vertical asymptotes at $-4,0$.