Limits of the form nonzero over zero

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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 $\boldsymbol {\tfrac {\#}{0}}$ ?

$\lim \limits _{x\to -1} \frac {1}{(x+1)^2}$ $\lim \limits _{x\to 2}\frac {x^2-3x+2}{x-2}$ $\lim \limits _{x\to 0}\frac {\sin (x)}{x}$ $\lim \limits _{x\to 2}\frac {x^2-3x-2}{x-2}$ $\lim \limits _{x\to 1}\frac {e^x}{\ln (x)}$

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, $\lim _{x\to -1} \frac {1}{(x+1)^2}\qquad \text {is of the form }\boldsymbol {\tfrac {\#}{0}}$ as $\lim _{x\to -1} 1 = 1 \qquad \text {and}\qquad \lim _{x\to -1}(x+1)^2 = 0.$ 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 positivenegative number.
The denominator is positivenegative 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 positivenegative number.
The denominator is positivenegative 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.