Vertical asymptotes

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}\hspace {0in}} \newcommand {\point }[1]{\left (#1\right )} \newcommand {\pt }[1]{\mathbf {#1}} \newcommand {\Lim }[2]{\lim _{\point {#1} \to \point {#2}}} \DeclareMathOperator {\proj }{\mathbf {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

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 $x=3$ .