Vertical asymptotes

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We explore functions that “shoot to infinity” near certain points.

Video Lecture

Consider the function $f(x) = \frac {1}{(x+1)^2}.$

While the $\lim _{x\rightarrow -1} f(x)$ does not exist, we learned in the last section that we can still say something about the behavior of the function.

1 : Which of the following are correct?

$\lim _{x\rightarrow -1} \frac {1}{(x+1)^2} = \infty$ $\lim _{x\rightarrow -1} \frac {1}{(x+1)^2} \rightarrow \infty$ $f(x) = \frac {1}{(x+1)^2}$ , so $f(-1) = \infty$ $f(x) = \frac {1}{(x+1)^2}$ , so as $x\rightarrow -1$ , $f(x) \rightarrow \infty$

The right arrow “ $\rightarrow$ ” means “goes to” or “approaches”. Remember that this is essentially what the limit is already saying; so saying $\lim _{x\rightarrow -1} \frac {1}{(x+1)^2} \rightarrow \infty$ is a little redundant. Indeed this would be translated to say “as x goes to -1, the function $\frac {1}{(x+1)^2}$ goes to ’goes to infinity’.” Which isn’t quite right. So if you use the “ $\lim$ ” notation, then you should use an equals sign. If you don’t, then you should use the “ $\rightarrow$ ” to signify that the function is approaching a value, but not “equal” to a value.

On the other hand, consider the function $f(x) = \frac {1}{(x-1)}.$

While the two sides of the limit as $x$ approaches $1$ do not agree, we can still consider the one-sided limits. We see $\lim _{x\rightarrow 1^+} f(x) = \infty$ and $\lim _{x\rightarrow 1^-} f(x) = -\infty$ .

If at least one of the following hold:

$\lim _{x\rightarrow a} f(x) = \pm \infty$ ,
$\lim _{x\rightarrow a^+} f(x) = \pm \infty$ ,
$\lim _{x\rightarrow a^-} f(x) = \pm \infty$ ,

then the line $x=a$ is a vertical asymptote of $f$ .

Find the vertical asymptotes of $f(x) = \frac {x^2-9x+14}{x^2-5x+6}.$

Since $f$ is a rational function, it is continuous on its domain. So the only points where the function can possibly have a vertical asymptote are zeros of the denominator.
Start by factoring both the numerator and the denominator: $\frac {x^2-9x+14}{x^2-5x+6} = \frac {(x-2)(x-7)}{(x-2)(x-3)}$ Using limits, we must investigate what happens with $f(x)$ when $x\rightarrow 2$ and $x\rightarrow 3$ , since $2$ and $3$ are the only zeros of the denominator. Write $\begin{align*} \lim_{x\rightarrow 2} \frac{\cancel{(x-2)}(x-7)}{\cancel{(x-2)}(x-3)} &= \lim_{x\rightarrow 2} \frac{(x-7)}{(x-3)}\\ &=\answer{5}. \end{align*}$

Now write

$\begin{align*} \lim_{x\rightarrow 3} \frac{\cancel{(x-2)}(x-7)}{\cancel{(x-2)}(x-3)} &= \lim_{x\rightarrow 3} \answer{\frac{(x-7)}{(x-3)}} \end{align*}$

Consider the one-sided limits separately.

When $x\rightarrow 3^+$ , the quantity $(x-3)$ is positive and approaches $0$ and the numerator is negative, therefore, $\lim _{x\rightarrow 3^+} f(x) = -\infty$ .

On the other hand, when $x\rightarrow 3^-$ , the quantity $(x-3)$ is negative and approaches $0$ and the numerator is negative, therefore, $\lim _{x\rightarrow 3^-} f(x) = \infty$ .

Hence we have a vertical asymptote at $x=3$ .