Vertical asymptotes

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We explore functions that ‘‘shoot to infinity’’ at certain points in their domain.

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

While the $\lim _{x\to -1} f(x)$ does not exist, something can still be said.

If $f(x)$ grows arbitrarily large as $x$ approaches $a$ , we write $\lim _{x\to a} f(x) = \infty$ and say that the limit of $f(x)$ approaches infinity as $x$ goes to $a$ .

If $|f(x)|$ grows arbitrarily large as $x$ approaches $a$ and $f(x)$ is negative, we write $\lim _{x\to a} f(x) = -\infty$ and say that the limit of $f(x)$ approaches negative infinity as $x$ goes to $a$ .

Which of the following are correct?

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 $\lim _{x\to 1^+} f(x) = \infty$ and $\lim _{x\to 1^-} f(x) = -\infty$ .

If at least one of the following hold:

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

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

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

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 when $x\to 2$ and $x\to 3$ . Write $\begin{align*} \lim_{x\to 2} \frac{(x-2)(x-7)}{(x-2)(x-3)} &= \lim_{x\to 2} \frac{(x-7)}{(x-3)}\\ &= \frac{-5}{-1}\\ &=5. \end{align*}$

Now write

$\begin{align*} \lim_{x\to 3} \frac{(x-2)(x-7)}{(x-2)(x-3)} &= \lim_{x\to 3} \frac{(x-7)}{(x-3)}\\ &= \lim_{x\to 3}\frac{-4}{x-3}. \end{align*}$

Consider the one-sided limits separately. Since $\lim _{x\to 3^+} (x-3)$ approaches $0$ from the right and the numerator is negative, $\lim _{x\to 3^+} f(x) = -\infty$ . Since $\lim _{x\to 3^-} (x-3)$ approaches $0$ from the left and the numerator is negative, $\lim _{x\to 3^-} f(x) = \infty$ .

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

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