3.5: Limits at Infinity
Learning Objectives
A student will be able to:
 Examine end behavior of functions on infinite intervals.
 Determine horizontal asymptotes.
 Examine indeterminate forms of limits of rational functions.
 Apply L’Hospital’s Rule to find limits.
 Examine infinite limits at infinity.
Introduction
In this lesson we will return to the topics of infinite limits and end behavior of functions and introduce a new method that we can use to determine limits that have indeterminate forms.
Examine End Behavior of Functions on Infinite Intervals
Suppose we are trying to analyze the end behavior of rational functions. Let's say we looked at some rational functions such as
We were then able to find infinite limits of more complicated rational functions such as
Now let’s consider other functions of the form
Example 1:
Consider the function
 Direct substitution leads to the indeterminate forms
00 and∞∞.  The function in the numerator is not a polynomial function, so we cannot use our previous methods such as applying
limx→∞1xp=0.
Let’s examine both the graph and values of the function for appropriate
We first note that domain of the function is
So,
Note: Please see Differentiation and Integration of Logarithmic and Exponential Functions in Chapter 6 for more on derivatives of Logarithmic functions.
So we infer that
For the infinite limit,
This unpredictable situation will apply to many other functions of the form. Hence we need another method that will provide a different tool for analyzing functions of the form
L’Hospital’s Rule: Let functions

limx→cf(x)g(x)=limx→cf′(x)g′(x) as long as this latter limit exists or is infinite.  If
f andg are differentiable at every numberx greater than some numbera , withg′(x)≠0, thenlimx→∞f(x)g(x)=limx→∞f′(x)g′(x) as long as this latter limit exists or is infinite.
Let’s look at applying the rule to some examples.
Example 2:
We will start by reconsidering the previous example,
Solution:
Since \begin{align*}\lim_{x\to 0} {\ln}(x + 1) = \lim_{x\to 0} x = 0\end{align*}
\begin{align*}\lim_{x\to 0} \frac{{\ln}(x + 1)} {x} = \lim_{x\to 0} \frac{\frac{1} {x + 1}} {1} = \frac{1} {1} = 1.\end{align*}
Likewise,
\begin{align*}\lim_{x \to \infty} \frac{{\ln}(x + 1)} {x} = \lim_{x \to \infty} \frac{\frac{1} {x + 1}} {1} = \frac{0} {1} = 0.\end{align*}
Now let’s look at some more examples.
Example 3:
Evaluate \begin{align*}\lim_{x\to 0} \frac{e^x  1} {x}.\end{align*}
Solution:
Since \begin{align*}\lim_{x\to 0} (e^x  1) = \lim_{x\to 0} x =0\end{align*}, L’Hospital’s Rule applies and we have
\begin{align*}\lim_{x\to 0} \frac{e^x  1} {x} = \lim_{x\to 0} \frac{e^x} {1} = \frac{1} {1} = 1.\end{align*}
Example 4:
Evaluate \begin{align*}\lim_{x \to +\infty} \frac{x^2} {e^x}\end{align*}
Solution:
Since \begin{align*}\lim_{x \to +\infty} x^2 = \lim_{x \to +\infty} e^x = + \infty\end{align*}, L’Hospital’s Rule applies and we have
\begin{align*}\lim_{x \to +\infty} \frac{x^2} {e^x} = \lim_{x \to +\infty} \frac{2x} {e^x}.\end{align*}
Here we observe that we still have the indeterminate form \begin{align*}\frac{\infty} {\infty}\end{align*}. So we apply L’Hospital’s Rule again to find the limit as follows:
\begin{align*}\lim_{x \to +\infty} \frac{x^2} {e^x} = \lim_{x \to +\infty} \frac{2x} {e^x} = \lim_{x \to +\infty} \frac{2} {e^x} = 0\end{align*}
L'Hospital's Rule can be used repeatedly on functions like this. It is often useful because polynomial functions can be reduced to a constant.
Let’s look at an example with trigonometric functions.
Example 5:
Evaluate \begin{align*}\lim_{x\to 0} \frac{1  \cos x}{x^2}.\end{align*}
Solution:
Since \begin{align*}\lim_{x\to 0} (1  \cos x) = \lim_{x\to 0} x^2 = 0\end{align*}, L’Hospital’s Rule applies and we have
\begin{align*}\lim_{x\to 0} \frac{1  \cos x} {x^2} = \lim_{x\to 0} \frac{\sin x} {2x} = \lim_{x\to 0} \frac{\cos x} {2} = \frac{1} {2}.\end{align*}
Lesson Summary
 We learned to examine end behavior of functions on infinite intervals.
 We determined horizontal asymptotes of rational functions.
 We examined indeterminate forms of limits of rational functions.
 We applied L’Hospital’s Rule to find limits of rational functions.
Multimedia Links
For an introduction to L'Hopital's Rule (8.0), see Khan Academy, L'Hopital's Rule (8:51).
Review Questions
 Use your graphing calculator to estimate \begin{align*}\lim_{x \to +\infty} \left [x[{\ln}(x + 3)  {\ln}(x)]\right].\end{align*}
 Use your graphing calculator to estimate \begin{align*}\lim_{x \to +\infty} \frac{x} {{\ln}(1 + 2e^x)}.\end{align*}
In problems #3–10, use L’Hospital’s Rule to compute the limits, if they exist.
 \begin{align*}\lim_{x\to 3} \frac{x^2 9} {x  3}\end{align*}
 \begin{align*}\lim_{x\to 0} \frac{\sqrt{1 + x}  \sqrt{1  x}} {x}\end{align*}
 \begin{align*}\lim_{x \to +\infty} \frac{{\ln}(x)} {\sqrt{x}}\end{align*}
 \begin{align*}\lim_{x \to +\infty} x^2 e^{2x}\end{align*}
 \begin{align*}\lim_{x\to 0} (1  x)^\frac{1}{x}\end{align*}
 \begin{align*}\lim_{x\to 0} \frac{e^x 1 x} {x^2}\end{align*}
 \begin{align*}\lim_{x \to \infty} \frac{e^x 1 x} {x^2}\end{align*}
 \begin{align*}\lim_{x \to \infty} x^{\frac{1} {4}} {\ln}(x)\end{align*}
Review Answers
 \begin{align*}\lim_{x \to +\infty} \left[x[{\ln}(x + 3)  {\ln} (x)]\right] = 3\end{align*}
 \begin{align*}\lim_{x \to +\infty} \frac{x} {{\ln}(1 + 2e^x)} = 1\end{align*}
 \begin{align*}\lim_{x\to 3} \frac{x^2 9} {x  3} = 6\end{align*}
 \begin{align*}\lim_{x\to 0} \frac{\sqrt{1 + x}  \sqrt{1  x}} {x} = 1\end{align*}
 \begin{align*}\lim_{x \to +\infty} \frac{{\ln} (x)} {\sqrt{x}} = 0\end{align*}
 \begin{align*}\lim_{x \to +\infty} x^2 e^{2x} = 0\end{align*}
 \begin{align*} \lim_{x\to 0} (1  x)^{\frac{1} {x}} = \frac{1} {e}\end{align*} : Hint: Let \begin{align*} (1  x)^{\frac{1} {x}} = e^{{\ln} (1  x)^{\frac{1} {x}}}\end{align*}, so \begin{align*} \lim_{x\to 0} (1  x)^{\frac{1} {x}} = \lim_{x \to 0} e^{{\ln}(1  x) \frac{1} {x}} = e^{\lim_{x \to 0} \frac{1} {x} {\ln}(1  x)}\end{align*}
 \begin{align*}\lim_{x\to 0} \frac{e^x 1 x} {x^2} = \frac{1}{2}\end{align*}
 \begin{align*}\lim_{x \to \infty} \frac{e^x 1 x} {x^2} = 0\end{align*}
 \begin{align*}\lim_{x \to \infty} x^{\frac{1} {4}} {\ln} (x) = 0\end{align*}