MA005: Calculus I

If our function $\mathrm{f}$ is the ratio of a polynomial $\mathrm{P}(\mathrm{x})$ and a polynomial $\mathrm{Q}(\mathrm{x}), \mathrm{f}(\mathrm{x})=\frac{\mathrm{P}(\mathrm{x})}{\mathrm{Q}(\mathrm{x})}$, then the only candidates for vertical asymptotes are the values of $x$ where $Q(x)=0$. However, the fact that $Q(a)=0$ is not enough to guarantee that the line $\mathrm{x}=\mathrm{a}$ is a vertical asymptote of $\mathrm{f}$; we also need to evaluate $\mathrm{P}(\mathrm{a})$. If $Q(a)=0$ and $P(a) \neq 0$, then the line $x=a$ is a vertical asymptote of $f$. If $Q(a)=0$ and $P(a)=0$, then the line $x=a$ may or may not be a vertical asymptote.

Example 7: Find the vertical asymptotes of $f(x)=\frac{x^{2}-x-6}{x^{2}-x}$ and $g(x)=\frac{x^{2}-3 x}{x^{2}-x}$.

Solution: $f(x)=\frac{x^{2}-x-6}{x^{2}-x}=\frac{(x-3)(x+2)}{x(x-1)}$ so the only values which make the denominator $0$ are $\mathrm{x}=0$ and $\mathrm{x}=1$, and these are the only candidates to be vertical asymptotes.

$\lim \limits_{x \rightarrow 0^{+}} f(x)=+\infty$ and $\lim \limits_{x \rightarrow 1^{+}} f(x)=-\infty$ so $x=0$ and $x=1$ are both vertical asymptotes of $f$

$g(x)=\frac{x^{2}-3 x}{x^{2}-x}=\frac{x(x-3)}{x(x-1)}$ so the only candidates to be vertical asymptotes are $x=0$ and $x=1$

$\lim \limits_{x \rightarrow 1^{+}} g(x)=\lim \limits_{x \rightarrow 1^{+}} \frac{x(x-3)}{x(x-1)}=\lim \limits_{x \rightarrow 1^{+}} \frac{x-3}{x-1}=-\infty$ so $x=1$ is a vertical asymptote of $g$

$\lim \limits_{x \rightarrow 0} g(x)=\lim \limits_{x \rightarrow 0} \frac{x(x-3)}{x(x-1)}=\lim \limits_{x \rightarrow 0} \frac{x-3}{x-1}=3 \neq \infty$ so $x=0$ is not a vertical asymptote.

Practice 6: Find the vertical asymptotes of $f(x)=\frac{x^{2}+x}{x^{2}+x-2}$ and $g(x)=\frac{x^{2}-1}{x-1}$.