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 \(\relax \boldsymbol {\tfrac {\#}{0}}\) 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 \(\relax \boldsymbol {\tfrac {\#}{0}}\)?
Let’s see what is going on with limits of the form \(\relax \boldsymbol {\tfrac {\#}{0}}\).
Consider the function
\[ f(x) = \frac {1}{(x+1)^2}. \]
Use a table of values to investigate \(\lim _{x\to -1} \frac {1}{(x+1)^2}\). Fill in the table below: ![[Picture]](digInLimitsOfTheFormNonZeroOverZero.online-076b7f11be0dafe565bb3baa56646b52.svg)
\[ \begin{array}{c|c|c} x & (x+1)^{2} & f(x) = \frac {1}{(x+1)^2}\\ \hline -1.1 & 0.01 & 100 \\ -1.01 & 0.0001 & 10000 \\ -1.001 & 0.000001 & \answer {1000000} \\ -1.0001 & 0.00000001 & \answer {100000000} \\ -0.9 & 0.01 & \answer {100} \\ -0.99 & 0.0001 & \answer {10000} \\ -0.999 & 0.000001 & \answer {1000000} \\ -0.9999 & 0.00000001 & \answer {100000000} \\ \end{array} \]
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 \(-1\) . Since \[ \lim _{x\to -1} 1 = 1 \qquad \text {and}\qquad \lim _{x\to -1}(x+1)^2 = 0 \]
we have that
\[ \lim _{x\to -1} \frac {1}{(x+1)^2}\qquad \text {is of the form }\numOverZero \]
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" does not mean that the limit exists. It
merely allows us to describe more precisely the behavior of the function \(f\) near \(a\), than
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 separately:
But, we can do better than that! As \(x\) approaches \(-2\):
\[ \lim _{x\to -2} e^x = \answer [given]{\frac {1}{e^2}}\qquad \text {and}\lim _{x\to -2}\left ((x+2)^4\right ) = 0. \]
So, this limit is of the
form \(\relax \boldsymbol {\tfrac {\#}{0}}\). 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
\[ \lim _{x\to -2} \frac {e^x}{(x+2)^4} = \infty . \]
In this example, when we say “As \(x\) approaches \(-2\)” we don’t mean that we’re checking a single value of \(x\). We mean ALL values of \(x\) that are close enough to \(-2\). Checking a single value is not enough to justify whether the numerator will be positive or negative for all \(x\) values sufficiently close to \(-2\). Constructing a sign-chart or graphing the numerator could be helpful with this determination.
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 separately.
\[ \lim _{x\to 3^+} \left (x^2-9x+14\right ) = \answer [given]{-4}\qquad \text {and}\lim _{x\to 3^+}\left (x^2-5x+6\right ) = 0 \]
This limit is of the form \(\relax \boldsymbol {\tfrac {\#}{0}}\). 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
\[ \lim _{x\to 3^+} \frac {x^2-9x+14}{x^2-5x+6} = -\infty . \]
Again, to determine if the numerator will be positive or negative for all \(x\) sufficiently close to, but greater than, \(3\), we cannot plug in a single value of \(x\). (Even plugging in \(x=3.0000001'\) would not be enough to justify our resullt.) We have to verify the sign for ALL values of \(x\) that are sufficiently close to, but greater than \(3\).
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 \(\relax \boldsymbol {\tfrac {\#}{0}}\). 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. We’re using \(\pm \infty \) to indicate directions, rather than a specific value.