Vertical asymptotes

$\newenvironment {prompt}{}{} \newcommand {\accordion }[0]{} \newcommand {\ungraded }[0]{} \newcommand {\xmDefaultPreamble }[0]{xmPreamble.tex} \newcommand {\xmDefaultPrintstyle }[0]{xmPrintstyle.sty} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{}$

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 $x=3$.

2025-01-06 19:32:17