Limits of the form nonzero over zero

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

What can be said about limits that have the form nonzero over zero?

Let’s cut to the chase:

A limit

$\lim _{x\to a} \frac {f(x)}{g(x)}$

is said to be of the form $\boldsymbol {\tfrac {\#}{0}}$ if

$\lim _{x\to a} f(x) = k\qquad \text {and}\qquad \lim _{x\to a} g(x) =0.$

where $k$ is some nonzero constant.

Which of the following limits are of the form $\boldsymbol {\tfrac {\#}{0}}$ ?

In our next example, let’s see what is going on with limits of the form $\boldsymbol {\tfrac {\#}{0}}$ .

Consider the function

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

Use a table of values to investigate $\lim \limits _{x\to -1} \frac {1}{(x+1)^2}$ .

Fill in the table below:

$\begin{array}{c|c|c} x & (x+1)^{2} & f(x) = \frac {1}{(x+1)^2}\\ \hline -1.1 & 0.01 & 100 \\ -1.01 & 0.0001 & 10000 \\ -1.001 & 0.000001 & \answer {1000000} \\ -1.0001 & 0.00000001 & \answer {100000000} \\ -0.9 & 0.01 & \answer {100} \\ -0.99 & 0.0001 & \answer {10000} \\ -0.999 & 0.000001 & \answer {1000000} \\ -0.9999 & 0.00000001 & \answer {100000000} \\ \end{array}$

What does the table tell us about

$\lim _{x\to -1} \frac {1}{(x+1)^2}?$

It appears that the limit does not exist, since the expression

$\frac {1}{(x+1)^2}$

becomes larger and larger as $x$ approaches $-1$ . So,

$\lim _{x\to -1} \frac {1}{(x+1)^2}\qquad \text {is of the form }\boldsymbol {\tfrac {\#}{0}}$

$\lim _{x\to -1} 1 = 1 \qquad \text {and}\qquad \lim _{x\to -1}(x+1)^2 = 0.$

Moreover, as $x$ approaches $-1$ :

The numerator is positive.
The denominator approaches zero and is positive.

Hence, the expression

$\frac {1}{(x+1)^2}$

will become arbitrarily large as $x$ approaches $-1$ . We can see this in the graph of $f$ .

We are now ready for our next definition.

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

Note: Saying “the limit is equal to infinity” describes more precisely the behavior of the function $f$ near $a$ , than just saying ”the limit does not exist”.

Let’s consider a few more examples.

Compute:

$\lim _{x\to -2} \frac {e^x}{(x+2)^4}$

First let’s look at the form of this limit. We do this by taking the limits of both the numerator and denominator:

$\lim _{x\to -2} e^x = \answer [given]{\frac {1}{e^2}}\qquad \text {and}\lim _{x\to -2}\left ((x+2)^4\right ) = 0.$

So, this limit is of the form $\boldsymbol {\tfrac {\#}{0}}$ . This form is determinate, since it implies that the limit does not exist.
But, we can do better than that! As $x$ approaches $-2$ :

The numerator is a
number.
The denominator is
and is approaching zero.

This means that

$\lim _{x\to -2} \frac {e^x}{(x+2)^4} = \infty .$

Compute:

$\lim _{x\to 3^+} \frac {x^2-9x+14}{x^2-5x+6}$

First let’s look at the form of this limit, which we do by taking the limits of both the numerator and denominator.

$\lim _{x\to 3^+} \left (x^2-9x+14\right ) = \answer [given]{-4}\qquad \text {and}\lim _{x\to 3^+}\left (x^2-5x+6\right ) = 0$

This limit is of the form $\boldsymbol {\tfrac {\#}{0}}$ . Next, we should factor the numerator and denominator to see if we can simplify the problem at all. $\begin{align*} \lim _{x\to 3^+}\frac {x^2-9x+14}{x^2-5x+6} &= \lim _{x\to 3^+}\frac {\cancel {(x-2)}(x-7)}{\cancel {(x-2)}(x-3)}\\ &= \lim _{x\to 3^+}\frac {x-7}{x-3} \end{align*}$

Canceling a factor of $x-2$ in the numerator and denominator means we can more easily check the behavior of this limit. As $x$ approaches $3$ from the right:

The numerator is a
number.
The denominator is
and approaching zero.

This means that

$\lim _{x\to 3^+} \frac {x^2-9x+14}{x^2-5x+6} = -\infty .$

Here is our final example.

Compute:

$\lim _{x\to 3} \frac {x^2-9x+14}{x^2-5x+6}$

We’ve already considered part of this example, but now we consider the two-sided limit. We already know that

$\lim _{x\to 3} \frac {x^2-9x+14}{x^2-5x+6} = \lim _{x\to 3}\frac {x-7}{x-3},$

and that this limit is of the form $\boldsymbol {\tfrac {\#}{0}}$ . We also know that as $x$ approaches $3$ from the right,

The numerator is a negative number.
The denominator is positive and approaching zero.

Hence our function is approaching $-\infty$ from the right.

As $x$ approaches $3$ from the left,

The numerator is negative.
The denominator is negative and approaching zero.

Hence our function is approaching $\infty$ from the left. This means

$\lim _{x\to 3} \frac {x^2-9x+14}{x^2-5x+6} = \answer [format=string,given]{DNE}.$

We can confirm our results of the previous two examples by looking at the graph of $y=\frac {x^2-9x+14}{x^2-5x+6}$ :

$\graph [xmin=-10,xmax=15,ymin=-10,ymax=10]{\frac {x^2-9x+14}{x^2-5x+6}}$

Some people worry that the mathematicians are passing into mysticism when we talk about infinity and negative infinity. However, when we write

$\lim _{x\to a} f(x) = \infty \qquad \text {and}\qquad \lim _{x\to a} g(x) = -\infty$

all we mean is that as $x$ approaches $a$ , $f(x)$ becomes arbitrarily large and $|g(x)|$ becomes arbitrarily large, with $g(x)$ taking negative values.

2025-01-06 19:54:11

Press...	...to do
left/right arrows	Move cursor
shift+left/right arrows	Select region
ctrl+a	Select all
ctrl+x/c/v	Cut/copy/paste
ctrl+z/y	Undo/redo
ctrl+left/right	Add entry to list or column to matrix
shift+ctrl+left/right	Add copy of current entry/column to to list/matrix
ctrl+up/down	Add row to matrix
shift+ctrl+up/down	Add copy of current row to matrix
ctrl+backspace	Delete current entry in list or column in matrix
ctrl+shift+backspace	Delete current row in matrix

Type...	...to get
norm	$\|\|\blue{[?]}\|\|$
text	$\text{\blue{[?]}}$
sym_name	$\backslash\texttt{\blue{[?]}}$
abs	$\left\|\blue{[?]}\right\|$
sqrt	$\sqrt{\blue{[?]}}$
paren	$\left(\blue{[?]}\right)$
floor	$\lfloor \blue{[?]} \rfloor$
factorial	$\blue{[?]}!$
exp	${\blue{[?]}}^{\blue{[?]}}$
sub	${\blue{[?]}}_{\blue{[?]}}$
frac	$\dfrac{\blue{[?]}}{\blue{[?]}}$
int	$\displaystyle\int{\blue{[?]}}d\blue{[?]}$
defi	$\displaystyle\int_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}d\blue{[?]}$
deriv	$\displaystyle\frac{d}{d\blue{[?]}}\blue{[?]}$
sum	$\displaystyle\sum_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}$
prod	$\displaystyle\prod_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}$
root	$\sqrt[\blue{[?]}]{\blue{[?]}}$
vec	$\left\langle \blue{[?]} \right\rangle$
mat	$\left(\begin{matrix} \blue{[?]} \end{matrix}\right)$
*	$\cdot$
infinity	$\infty$
arcsin	$\arcsin\left(\blue{[?]}\right)$
arccos	$\arccos\left(\blue{[?]}\right)$
arctan	$\arctan\left(\blue{[?]}\right)$
sin	$\sin\left(\blue{[?]}\right)$
cos	$\cos\left(\blue{[?]}\right)$
tan	$\tan\left(\blue{[?]}\right)$
sec	$\sec\left(\blue{[?]}\right)$
csc	$\csc\left(\blue{[?]}\right)$
cot	$\cot\left(\blue{[?]}\right)$
log	$\log\left(\blue{[?]}\right)$
ln	$\ln\left(\blue{[?]}\right)$
alpha	$\alpha$
beta	$\beta$
gamma	$\gamma$
delta	$\delta$
epsilon	$\epsilon$
zeta	$\zeta$
eta	$\eta$
theta	$\theta$
iota	$\iota$
kappa	$\kappa$
lambda	$\lambda$
mu	$\mu$
nu	$\nu$
xi	$\xi$
omicron	$\omicron$
pi	$\pi$
rho	$\rho$
sigma	$\sigma$
tau	$\tau$
upsilon	$\upsilon$
phi	$\phi$
chi	$\chi$
psi	$\psi$
omega	$\omega$
Gamma	$\Gamma$
Delta	$\Delta$
Theta	$\Theta$
Lambda	$\Lambda$
Xi	$\Xi$
Pi	$\Pi$
Sigma	$\Sigma$
Phi	$\Phi$
Psi	$\Psi$
Omega	$\Omega$

Controls

Symbols

Settings