Vertical asymptotes

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

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 $f(x)$ is equal to infinity as $x$ goes to $a$ .

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

Which of the following are correct?

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

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 $f(x) = \frac {x^2-9x+14}{x^2-5x+6}.$

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 $x\to 2$ and $x\to 3$ , since $2$ and $3$ are the only zeros of the denominator. Write $\begin{align*} \lim_{x\to 2} \frac{\cancel{(x-2)}(x-7)}{\cancel{(x-2)}(x-3)} &= \lim_{x\to 2} \frac{(x-7)}{(x-3)}\\ &= \frac{-5}{-1}\\ &=5. \end{align*}$ Now write $\begin{align*} \lim_{x\to 3} \frac{\cancel{(x-2)}(x-7)}{\cancel{(x-2)}(x-3)} &= \lim_{x\to 3} \frac{(x-7)}{(x-3)}\\ &= \lim_{x\to 3}\frac{-4}{x-3}. \end{align*}$ Consider the one-sided limits separately.

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

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

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