### Motivating Questions

- How can we evaluate complicated limits as \(x \to \infty\text{?}\)
- How can we evaluate complicated limits as \(x \to -\infty\text{?}\)

- How can we evaluate complicated limits as \(x \to \infty\text{?}\)
- How can we evaluate complicated limits as \(x \to -\infty\text{?}\)

In this section we continue with the general goal of Section 3.6: evaluating limits. In this section we will focus on limits at infinity, which have the form

\begin{equation*}
\lim_{x \to \infty} f(x) \text{ and } \lim_{x \to -\infty} f(x)\text{.}
\end{equation*}

These are important for determining the *end behavior* of a function, which can tell us how a system behaves in the long run. They can help us answer questions like “is the population headed towards extinction?” and “what is the baseline concentration of a substance in the bloodstream?”. We were introduced to basic limits of this form in Subsection 3.6.2. In this section, we will use the *method of leading behaviors* to help us evaluate more complicated limits at infinity.

- Graph \(y=x^4 - 2x + \sqrt{x} -10\) and \(y=x^4\) in Desmos. Compare and contrast the graphs that you see. What is the same? What is different?
- Graph \(y=-3x^5 + 10x + \ln(x) -50\) and \(y=-3x^5\) in Desmos. Compare and contrast the graphs that you see. What is the same? What is different?
- Graph \(g(x)=x^{10} + 2x + \sqrt{x} -e^x\) in Desmos. What is \(\lim_{x \to \infty} g(x)\text{?}\) Why do you think this is the limit’s value, and how could you determine this value just using the equation of \(g\text{?}\)

In Warm-Up 3.7.1 we observed a complicated function built as the sum of simpler functions can exhibit *local behavior* that is complicated to predict without computation, like its zeros and its local extrema. However, if we are interested in describing the *end behavior* of such a function, as we are in this section, it can be predicted based on the behavior of a single simpler function. Indeed, if \(f(x) = f_1(x) + f_2(x) + f_3(x)\text{,}\) and \(|f_1(x)|\) dominates \(|f_2(x)|\) and \(|f_3(x)|\) as \(x \to \infty\text{,}\) then the graph of \(f\) will start to behave like the graph of \(f_1\) as \(x \to \infty\text{.}\) For such a function, we give the dominant term a special name.

Let \(f(x)\) be a sum of functions. If one function in the sum dominates all of the other functions in the sum in absolute value as \(x \to \infty\text{,}\) we call this function the leading behavior of \(f\) at \(\infty\). We denote the leading behavior of \(f\) at \(\infty\) as \(f_\infty(x)\text{.}\)

Note that leading behavior is determined by which term dominates *in absolute value*. A term being positive or negative can impact the value of the limit, but it does not effect whether or not a term is the leading behavior. As we described above, the leading behavior of a function at \(\infty\) can be useful in evaluating limits at infinity:

Let \(f(x)\) be a function with leading behavior at \(\infty\) \(f_\infty(x)\text{.}\) Then

\begin{equation*}
\lim_{x \to \infty} f(x) = \lim_{x \to \infty} f_\infty(x)\text{.}
\end{equation*}

The benefit of leading behaviors is that it allows us to replace a complicated function with a simpler one that has the same end behavior.

Let \(g(x) = -x^3 - \sqrt{x} + \ln(x)\) and \(h(x)= e^{-x} +2^x + x^{100}\text{.}\) In Section 3.6 we determined that as \(x \to \infty\text{,}\) exponential growth dominates power growth dominates logarithmic growth. Therefore, we know that \(g_\infty(x) = -x^3\text{,}\) since \(|-x^3|\) dominates the other terms as \(x \to \infty\text{.}\) Also, \(h_\infty(x) = 2^x\) since exponential growth dominates power function growth. We must be careful to notice here that the leading behavior at infinity is not \(e^{-x}\text{,}\) since this function decays to \(0\) as \(x \to \infty\text{.}\)

We could also use leading behaviors to evaluate the limit of the ratio of \(g\) and \(h\text{:}\)

\begin{equation*}
\lim_{x \to \infty} \dfrac{-x^3 - \sqrt{x} + \ln(x)}{e^{-x} +2^x + x^{100}} = \lim_{x \to \infty} \dfrac{g_\infty(x)}{h_\infty(x)} = \lim_{x \to \infty} \dfrac{-x^3}{2^x} = 0\text{,}
\end{equation*}

since \(2^x\) dominates \(|-x^3|\) as \(x \to \infty\text{.}\)

For each function pair \(f,g\) below, first determine \(f_\infty(x)\) and \(g_\infty(x)\text{.}\) Then evaluate \(\lim_{x \to \infty} \dfrac{f(x)}{g(x)}\text{.}\)

- \(f(x) = e^{-2x} + x + e^{0.1x}\text{,}\) \(g(x)= x^2 -x + \ln(x)\)
- \(f(x) = x^{-10}+ \sqrt{x} + 0.5^x\text{,}\) \(g(x)= -\ln(x) +x^{-1} \)
- \(f(x) = -3x^2+ \sqrt[3]{x^5} + 0.5^x\text{,}\) \(g(x)= x^2 +x^{-1} \)

Leading behavior at \(-\infty\) is defined similarly to leading behavior at \(\infty\text{:}\) ####
Definition 3.7.3.

Let \(f(x)\) be a sum of functions. If one function in the sum dominates all of the other functions in the sum in absolute value as \(x \to -\infty\text{,}\) we call this function the leading behavior of \(f\) at \(-\infty\). We denote the leading behavior of \(f\) at \(-\infty\) as \(f_{-\infty}(x)\text{.}\)

We can also use \(f_{-\infty}(x)\) similarly to help us evaluate limit as \(x \to -\infty\text{:}\) #### Leading Behavior at \(-\infty\).

The main thing we must be careful of with leading behaviors at \(-\infty\) is to pay attention to the behavior of the function as \(x \to -\infty\text{;}\) that is, as we move to the left on the graph of the function. This behavior is in general different than the behavior as \(x \to \infty\) (as we move to the right on the graph of the function). For example, \(e^x\) is a common leading behavior at \(\infty\) because it is exponential growth, and gets large quickly as \(x \to \infty\text{.}\) However, it is not a common leading behavior at \(-\infty\) since it decays to \(0\) as \(x \to -\infty\text{:}\) \(\lim_{x \to -\infty} e^x = 0\text{.}\)

Let \(f(x)\) be a function with leading behavior at \(-\infty\) \(f_{-\infty}(x)\text{.}\) Then

\begin{equation*}
\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} f_{-\infty}(x)\text{.}
\end{equation*}

Let \(g(x) = -x^3 - \sqrt[3]{x} + \ln(-x)\) and \(h(x)= e^{-x} +2^x + x^{100}\text{.}\) Then \(g_{-\infty}(x) = -x^3\text{,}\) since \(|-x^3|\) dominates the other terms as \(x \to -\infty\text{.}\) For \(h(x)\text{,}\) note that \(2^x\) gets very small as \(x \to -\infty\text{.}\) The function \(x^{100}\) gets very large as \(x \to -\infty\text{,}\) but is power function growth. The function \(e^{-x}\) grows exponentially as \(x \to -\infty\text{.}\) Thus, \(h_{-\infty}(x) = e^{-x}\text{.}\)

We can use leading behaviors to evaluate the limit of the ratio of \(g\) and \(h\text{:}\)

\begin{equation*}
\lim_{x \to -\infty} \dfrac{-x^3 - \sqrt[3]{x} + \ln(-x)}{e^{-x} +2^x + x^{100}} = \lim_{x \to -\infty} \dfrac{g_{-\infty}(x)}{h_{-\infty}(x)} = \lim_{x \to -\infty} \dfrac{-x^3}{e^{-x}} = 0\text{,}
\end{equation*}

since \(e^{-x}\) dominates \(|-x^3|\) as \(x \to -\infty\text{.}\)

For each function pair \(f,g\) below, first determine \(f_{-\infty}(x)\) and \(g_{-\infty}(x)\text{.}\) Then evaluate \(\lim_{x \to -\infty} \dfrac{f(x)}{g(x)}\text{.}\)

- \(f(x) = e^{-2x} + x + e^{0.1x}\text{,}\) \(g(x)= x^2 -x + \ln(-x)\)
- \(f(x) = x^{-10}+ \sqrt{-x} + 0.5^x\text{,}\) \(g(x)= -\ln(-x) +x^{-1} \)
- \(f(x) = -3x^2+ \sqrt[3]{x^5} + 0.5^x\text{,}\) \(g(x)= x^2 +x^{-1} \)

Answer.#### Question 3.7.5.

How can we evaluate complicated limits as \(x \to \infty\text{?}\)To evaluate limits of the form \(\lim_{x\to \infty} f(x)\text{,}\) we can determine if \(f(x)\) has a leading behavior at \(\infty\text{,}\) \(f_\infty(x)\text{.}\) If so, we know\begin{equation*} \lim_{x\to \infty} f(x) = \lim_{x\to \infty} f_\infty(x)\text{.} \end{equation*} Answer.#### Question 3.7.6.

How can we evaluate complicated limits as \(x \to -\infty\text{?}\)To evaluate limits of the form \(\lim_{x\to -\infty} f(x)\text{,}\) we can determine if \(f(x)\) has a leading behavior at \(-\infty\text{,}\) \(f_{-\infty}(x)\text{.}\) If so, we know\begin{equation*} \lim_{x\to -\infty} f(x) = \lim_{x\to -\infty} f_{-\infty}(x)\text{.} \end{equation*}

- Determine \(k_\infty(x)\text{.}\)
- Evaluate \(\displaystyle \lim_{x \to \infty}\frac{e^x}{e^x + e^{-x}}\text{.}\)
- Determine \(k_{-\infty}(x)\text{.}\)
- Evaluate \(\displaystyle \lim_{x \to -\infty}\frac{e^x}{e^x + e^{-x}}\text{.}\)

- A function \(h\) such that \(\lim_{x \to \infty} h(x) = 3\) and \(\lim_{x \to -\infty} h(x) =0\text{.}\)
- A function \(j\) such that \(\lim_{x \to \infty} j(x) = 3\) and \(\lim_{x \to -\infty} j(x) =2\text{.}\)