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. 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: ![[Picture]](digInLimitsOfTheFormNonZeroOverZero-7961de85d617e67e235a1a0f35a1d0f5.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 \(-1\) . 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:
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 \(\boldsymbol {\tfrac {\#}{0}}\). 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.
2025-01-06 19:54:11