Vertical asymptotes

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

Let’s return to the function $f(x) = \frac {1}{(x+1)^2}.$

As we stated learned on the previous card, $\lim _{x\to -1} f(x)$ does not exist, but something can still be said.

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:

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

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 $\begin{align*} \displaystyle\lim_{x\to 2} \frac{(x-2)(x-7)}{(x-2)(x-3)} &= \lim_{x\to 2} \frac{(x-7)}{(x-3)}\\ &= \frac{-5}{-1}\\ &=5. \end{align*}$ 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 $\displaystyle \lim _{x\to 3^+} (x-3)$ approaches $0$ from the right and the numerator is negative, $\displaystyle \lim _{x\to 3^+} f(x) = -\infty$ . Already, this is enough to conclude that we have a vertical asymptote at $x=3$ . Since $\displaystyle \lim _{x\to 3^-} (x-3)$ approaches $0$ from the left and the numerator is negative, $\displaystyle \lim _{x\to 3^-} f(x) = \infty$ .

From the one-sided limit information, we can conclude that $\displaystyle \lim _{x\to 3} f(x)$ does not exist. Notice that despite the fact that this limit does not exist, $f(x)$ still has a vertical asymptote at $x=3$ !

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