Limits At Infinity, Part II

Exponential Functions

In this section we’re going to review one of the more common functions in both calculus and the sciences. However, before getting to this function let’s take a much more general approach to things.

Let’s start with $b\gt 0, b\neq 1$. An exponential function is then a function in the form, $$f(x) = b^x.$$

Note that we avoid $b=1$ because that would give the constant function, $f(x)=1$, and we avoid $b=0$ since this would also give a constant function. We avoid negative values of $b$ for the following reason.

Let’s, for a second, suppose that we did allow $b$ to be negative and look at the following function.

Let's do some evaluation.

So, for some values of $x$ we will get real numbers and for other values of $x$ we will get complex numbers. We want to avoid this, so if we require $b\gt 0$, this will not be a problem.

Let’s take a look at a couple of exponential functions.

Sketch the graphs of $f(x) = 2^x$ and $g(x) = \left(\frac12\right)^x$.

Click for solution

When in doubt, we can create a table of values. We'll work on $f(x) = 2^x$ first.

$$\begin{array}{c|c} x & f(x) = 2^x\\ \hline -2 & 2^{-2} = \frac{1}{2^2} = \frac14\\ -1 & 2^{-1} = \frac{1}{2^1} = \frac12\\ 0 & 2^0 = 1\\ 1 & 2^1 = 2\\ 2 & 2^2 = 4\\ \end{array}$$

We can sketch these points and connect the dots with a smooth line to reveal the classic exponential growth shape:

Now, let's investigate $g(x) = \left(\frac12\right)^x$ with its own table of values.

$$\begin{array}{c|c} x & g(x) = \left(\frac12\right)^x\\ \hline -2 & \left(\frac12\right)^{-2} = \frac{1^{-2}}{2^{-2}} = \frac{2^2}{1^2} = 4\\ -1 & \left(\frac12\right)^{-1} = \frac{1^{-1}}{2^{-1}} = \frac{2}{1} = 2\\ 0 & \left(\frac12\right)^0 = 1\\ 1 & \left(\frac12\right)^1 = \frac12\\ 2 & \left(\frac12\right)^2 = \frac14\\ \end{array}$$

Again, we'll sketch these ordered pairs and connect the dots with a smooth line to see exponential decay:

These graphs illustrates some very nice properties about exponential functions in general.

But why are we allowed to connect those dots?

Note: The following content is adapted from OpenStax and is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction.

When we create those graphs, we're working off of an assumption that $2^x$ is defined for any real number that we could choose to substitute in for $x$. Why should that be true, though?

When we first learn about exponential functions, we think of what happens with $b^x$ when $x$ is a positive integer (a positive whole number). That is, $b^x$ is the product of $b$ multiplied by itself $n$ times. Under that definition, we can start to establish a pattern:

As our $x$-values decrease by 1, our $y$-values are divided by a factor of $b$. This is why we end up with the properties that $b^0 = 1$; it must follow from that pattern. This is also why a negative integer "flips" the position of $b$ over the fraction bar. We must again divide by $b$, so we have $\frac{1}{b}$.

Part of our beginning notions of exponential functions also include the following:

These ideas allow us to evaluate $b^x$ for any rational (fraction) value of $x$, but what about other real numbers? For instance, what is the value of $4^\pi$? The number $\pi$ is irrational, so it cannot be expressed as a fraction, so we cannot use any of these other ideas of how to evaluate $4^\pi$, but we'd like to be able to find a value for it. What we can do is set up a sequence of relationships.

We know that the number $\pi \approx 3.14159265...$ is between 3.1 and 3.2, so $4^{3.1}\lt 4^\pi \lt 4^{3.2}$. What if we want to get a little closer to the $4^\pi$ in the middle? We could then say that $\pi$ is between 3.14 and 3.15, so $4^{3.14} \lt 4^\pi \lt 4^{3.15}$ to get a more precise idea of what that value of $4^\pi$ might be. In fact, we can create a little sequence that gets closer and closer to sandwiching in that $4^\pi$ with each iteration to create a slightly better approximation each time: $$\begin{array}{rcl} 4^3 = 64 &\lt ~~4^\pi &\lt 4^4 = 256\\ 4^{3.1}\approx 73.5166947198 &\lt~~ 4^\pi &\lt 4^{3.2}\approx 84.4485062895\\ 4^{3.14} \approx 77.7084726013 &\lt ~~4^\pi &\lt 4^{3.15}\approx 78.7932424541\\ 4^{3.141}\approx 77.8162741237 &\lt ~~4^\pi &\lt 4^{3.142}\approx 77.9242251944\\ 4^{3.1415}\approx 77.8702309526 &\lt ~~4^\pi &\lt 4^{3.1416}\approx 77.9242251944\\ 4^{3.14159} \approx 77.8799471543 &\lt~~ 4^\pi &\lt 4^{3.141593} \approx 77.8802710486\\ \end{array}$$ So, we can see that $4^\pi \approx 77.88$.

It is possible to show, though it is beyond the scope of this course, that $4^\pi$ is the only number that is greater than all of this pattern: $4^3, 4^{3.1}, 4^{3.14}, 4^{3.141}, 4^{3.1415}, ...$ and less than all of this pattern: $4^4, 4^{3.2}, 4^{3.15}, 4^{3.142}, 4^{3,1416}, ...$. The reason then that we can take a table like this: $$\begin{array}{c|c} x & 4^x\\ \hline -2 & \frac{1}{16}\\ -1 & \frac{1}{4}\\ 0 & 1\\ 1 & 4\\ 2 & 16\\ \end{array}$$ and connect those ordered pairs to make a continuous graph of $4^x$ is by allowing ourselves to define $b^r$ for any real value of $r$ by using similar patterns of approximation. In general, we can say

Note also that exponential functions satisfy the general laws of exponents. To remind you of these laws, we state them as rules.

In the spirit of a thorough review, let's practice applying these laws to simplify expressions.

The number e

A special type of exponential function appears frequently in real-world applications. To describe it, consider the following example of exponential growth, which arises from compounding interest in a savings account. Suppose a person invests $P$ dollars in a savings account with an annual interest rate $r$, compounded annually. This means that after a year, we'll calculate the interest and add it in to the account. The next year, then, we'll calculated the interest based on the total amount in the account (the original principal plust the interest).

The amount of money after 1 year is $$A(1) = P + rP = P(1+r).$$

The amount of money after 2 year is $$A(2) = A(1)+rA(1) = P(1+r)+rP(1+r) = P(1+r)^2.$$

More generally, the amount after $t$ years is $$A(t)= P(1+r)^t.$$

Interest rates are listed on an annual basis, so if the money is compounded 2 times per year, we'll only add in half of the stated interest rate each time. Therefore, the amount of money after half a year is $$A\left(\frac12\right) = P+\left(\frac{r}{2}\right)P = P\left(1+\frac{r}{2}\right).$$

The amount of money after 1 year is $$A(1) = A\left(\frac12\right) + \left(\frac{r}{2}\right)A\left(\frac12\right)$$ $$= P\left(1+\frac{r}{2}\right)+\frac{r}{2}\left(P\left(1+\frac{r}{2}\right)\right) = P\left(1+\frac{r}{2}\right)^2.$$

After $t$ years, the amount of money in the account is $$A(t) = P\left(1+\frac{r}{2}\right)^{2t}.$$

More generally, if the money is compounded $n$ times per year, the amount of money in the account after $t$ years is given by the function $$A(t) = P\left(1+\frac{r}{n}\right)^{nt}.$$

Now, let's look at what happens to this quantity if we let $r = 1$ (that's a 100\% interest rate), let $t = 1$ (we'll only look at one year), let $P = 1$ (we'll invest one dollar), and let $n$ become very large. Letting $n$ go to infinity in this manner is called continuously compounding interest. Here is what we find:

Investigating the table, it looks like we have $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n \approx 2.718282.$$

This turns out to be a special, irrational number, named e.

The letter $e$ was first used to represent this number by the Swiss mathematician Leonhard Euler during the 1720s. Although Euler did not discover the number, he showed many important connections between $e$ and logarithmic functions. We still use the notation $e$ today to honor Euler’s work because it appears in many areas of mathematics and because we can use it in many practical applications.

Returning to our savings account example, we can conclude that if a person puts $P$ dollars in an account at an annual interest rate $r$, compounded continuously, then $A(t)=Pe^{rt}$. This function may be familiar. Since functions involving base $e$ arise often in applications, we call the function $f(x)=e^x$ the natural exponential function. Not only is this function interesting because of the definition of the number $e$, but also, as discussed next, its graph has an important property.

Since $e\gt 1$, we know $e^x$ is increasing on $(-\infty, \infty)$. In the graph below, we show a graph of $f(x)=e^x$ along with a tangent line to the graph of at $x=0$. We give a precise definition of tangent line in the next chapter; but, informally, we say a tangent line to a graph of $f$ at $x=a$ is a line that passes through the point $(a,f(a))$ and has the same “slope” as $f$ at that point.

The function $f(x)=e^x$ is the only exponential function $b^x$ with tangent line at $x=0$ that has a slope of 1. As we see later in the text, having this property makes the natural exponential function the most simple exponential function to use in many instances.

Limits of Exponential Functions

Note: The following content is adapted from Paul's Online Math Notes, with permission, and is not shared under the Creative Commons license.

Now that we've reviewed and filled in some concepts on how exponential functions work, let's examine the limits of exponential functions.

Evaluate each of the following limits

$$\lim_{x\to \infty}e^x~~~~\lim_{x\to-\infty}e^x$$ $$\lim_{x\to \infty}e^{-x}~~~~\lim_{x\to-\infty}e^{-x}$$ Click for solution

For these limits, we start by relying on the our knowledge of the graphs of exponential functions. Since $e\gt 1$, its graph will exhibit exponential growth:

Therefore, $$\lim_{x\to \infty}e^x = \infty ~~~\text{ and }~~~\lim_{x\to -\infty} e^x = 0.$$

Now, what to do about $e^{-x}$? Let's suppose we forgot how to create a graph for $e^{-x}$. One thing we can do is a bit of substitution. Let's let $t = -x$. Then as $x \to \infty$, $t \to -\infty$, so we can rewrite our next limit as such: $$\lim_{\textcolor{blue}{x\to \infty}}e^{\textcolor{purple}{-x}} \Rightarrow \lim_{\textcolor{blue}{t\to-\infty}}e^{\textcolor{purple}{t}}.$$ The function $e^t$ will still exhibit exponential growth, so $\lim\limits_{t\to-\infty}e^t = 0$, therefore the original limit is $$\lim_{x\to \infty}e^{-x} = 0.$$

For the sake of practice, we'll take the same approach with our final limit. If we let $t = -x$, then as $x\to -\infty$, $t\to \infty$, so $$\lim_{\textcolor{blue}{x\to-\infty}}e^{\textcolor{purple}{-x}} \Rightarrow \lim_{\textcolor{blue}{t\to\infty}}e^{\textcolor{purple}{t}} = \infty.$$

As we hinted, though, we can reach these same conclusions if we remember how graphical transformations work. The graph of $e^{-x}$ is the result of reflecting the graph of $e^{x}$ over the y-axis, so the graph of $e^{-x}$ looks something like this:

and therefore we still conclude that $$\lim_{x\to\infty}e^{-x} = 0 ~~~\text{ and }~~~\lim_{x\to-\infty}e^{-x} = \infty.$$

Here’s a quick set of examples to illustrate these ideas.

Let’s work some more complicated examples involving exponentials. In the following set of examples it won’t be that the exponents are more complicated, but instead that there will be more than one exponential function to deal with.

Let’s next take a look at some rational functions involving exponentials.

Logarithmic Functions

Once more, we'll take a break, this tim to look at a function that is related to the exponential functions: logarithms. Logarithms are one of the functions that students fear the most. The main reason for this seems to be that they simply have never really had to work with them. Once they start working with them, students come to realize that they aren’t as bad as they first thought.

We can investigate this relationship using some quick evaluations.

There are a couple of special logarithms that arise in many places. These are,

Let’s take a look at a couple of more logarithm evaluations, some of which deal with the natural or common logarithm and some of which don’t.

This last set of examples leads us to some of the basic properties of logarithms.

The last two properties will be especially useful in the next section. Notice as well that these last two properties tell us that, $$f(x) = b^x ~~~\text{and}~~~g(x) = \log_b(x)$$ are inverses of each other.

Here are some more properties that are useful in the manipulation of logarithms.

Note that there is no equivalent property to the first two for sums and differences. In other words, $$\log_b(x+y) \text{ IS NOT THE SAME AS } \log_b(x)+\log_b(y).$$ $$\log_b(x-y) \text{ IS NOT THE SAME AS } \log_b(x)-\log_b(y).$$

The last topic that we need to look at in this section is the change of base formula for logarithms. The change of base formula is,

This is the most general change of base formula and will convert from base $b$ to base $a$. However, the usual reason for using the change of base formula is to compute the value of a logarithm that is in a base that you can’t easily deal with. Using the change of base formula means that you can write the logarithm in terms of a logarithm that you can deal with. The two most common change of base formulas are

In fact, often you will see one or the other listed as THE change of base formula!

In the first part of this section we computed the value of a few logarithms, but we could do these without the change of base formula because all the arguments could be written in terms of the base to a power. For instance,

However, this only works because 49 can be written as a power of 7. We would need the change of base formula to compute $\log_7(50)$. $$\log_7(50) = \frac{\ln(50)}{\ln(7)} \approx \frac{3.91202300543}{1.94591014906} \approx 2.0103821378$$ OR $$ \log_7(50) = \frac{\log(50)}{\log(7)} \approx \frac{1.69897000434}{0.845098040014} \approx 2.0103821378$$

So, it doesn’t matter which we use, we will get the same answer regardless of the logarithm that we use in the change of base formula.

Graphs and limits of logarithmic functions

In order to see how the graph of a logarithmic function behaves, let's start by comparing it to an exponential function. We'll begin with tables of values:

Notice how the columns of inputs and outputs are flip-flopped? That's a hallmark of inverse functions. The way that relationship shows up graphically is that the graph of $\log_2(x)$ will be a reflection of the graph of $2^x$ about the line $y = x$.

What does this relationship tell us about limits, then? Well, we see on the left side of the exponential graph that $$\lim_{x\to\textcolor{blue}{-\infty}}(2^x) = \textcolor{purple}{0^+},$$ and so on the logarithmic graph, we have $$\lim_{x\to \textcolor{purple}{0^+}} (\log_2(x)) = \textcolor{blue}{-\infty}.$$ Similarly, $$\lim_{x\to\textcolor{blue}{\infty}}(2^x) = \textcolor{purple}{\infty}$$ and therefore $$\lim_{x\to \textcolor{purple}{\infty}}(\log_2(x)) = \textcolor{blue}{\infty}.$$

Evaluate each of the following limits

$$\lim_{x\to0^+}\ln(x)~~~~~~~~~\lim_{x\to\infty}\ln(x)$$ Click for solution

We'll begin by recalling that $\ln(x)$ is the same as $\log_e(x)$. That means the graph of $\ln(x)$ should be the inverse of the graph of $e^x$. The number $e$ is greater than 1, so we expect these graphs to look like this:

Looking at the graph of $\ln(x)$, then, we see that $$\lim_{x\to0^+}\ln(x) = -\infty~~~~~~~~~\lim_{x\to\infty}\ln(x) = \infty.$$

Note that we had to do a right-handed limit for the first one since we can’t plug negative $x$’s into a logarithm. This means that the normal limit won’t exist since we must look at $x$’s from both sides of the point in question and $x$’s to the left of zero are negative.

Let’s take a quick look at some logarithm examples.

Arctangent

As a final set of examples let’s take a look at some limits involving inverse tangents.

For a quick review, arctangent (denoted $\arctan(x)$ or $\tan^{-1}(x)$) is the inverse of the tangent function, and so it is the result of reflecting the tangent graph about the line $y=x$ (just like the relationship between exponential and logarithmic functions). We also need to limit the domain of the tangent function before we reflect it, so we don't end up with a weird graph that is no longer a function. We'll talk more about that in later sections, or you can check out this section of our trig textbook.

For our current purposes, here is the graph of arctangent:

With that in mind, let's try some limits.

We can use the results of the last two examples to help us with some more complicated limits involving the inverse tangent function. Let's try a few.

Practice Problems

For the following problems, evaluate (a) $\lim\limits_{x\to-\infty} f(x)$ and (b) $\lim\limits_{x\to \infty} f(x)$.

$f(x) = e^{8+2x-x^3}$
Click for solution
1. Let $t = 8+2x-x^3$. We know that $\lim\limits_{x\to-\infty}(8+2x-x^3) = \infty$, so as $x\to-\infty$, we have $t\to \infty$. Therefore $\lim\limits_{x\to-\infty}e^{8+2x-x^3} = \lim\limits_{t\to\infty}e^t = \infty$.
2. Let $t = 8+2x-x^3$. We know that $\lim\limits_{x\to\infty}(8+2x-x^3) = -\infty$, so as $x\to\infty$, we have $t\to -\infty$. Therefore $\lim\limits_{x\to\infty}e^{8+2x-x^3} = \lim\limits_{t\to-\infty}e^t = 0$.



$f(x) = e^{\left(\frac{6x^2+x}{5+3x}\right)}$
Click for solution
1. Let $t = \frac{6x^2+x}{5+3x}$. We can factor out the highest power of $x$ from the denominator to find the limit as $x\to-\infty$: $$\lim_{x\to-\infty}\frac{6x^2+x}{5+3x} = \lim_{x\to-\infty}\frac{x(6x+1)}{x(\frac{5}{x}+3)}$$ $$=\lim_{x\to-\infty}\frac{6x+1}{\frac{5}{x}+3} = \frac{-\infty}{0+3} = -\infty.$$ This tells us that as $x\to-\infty$, $t\to\infty$ as well. So our limit can be rewritten and evaluated as such: $$\lim_{x\to-\infty}e^{\left(\frac{6x^2+x}{5+3x}\right)} = \lim_{t\to-\infty}e^t = 0.$$
2. We will again let $t = \frac{6x^2+x}{5+3x}$ and factor out the highest power of $x$ from the denominator to find the limit as $x\to \infty$: $$\lim_{x\to\infty}\frac{6x^2+x}{5+3x} = \lim_{x\to\infty}\frac{x(6x+1)}{x(\frac{5}{x}+3)}$$ $$=\lim_{x\to\infty}\frac{6x+1}{\frac{5}{x}+3} = \frac{\infty}{0+3} = \infty.$$ Therefore, as $x\to\infty$, our $t\to\infty$, so our limit becomes: $$\lim_{x\to\infty}e^{\left(\frac{6x^2+x}{5+3x}\right)} = \lim_{t\to\infty}e^t = \infty.$$



$f(x) = 2e^{6x}-e^{-7x}-10e^{4x}$
Click for solution
1. We'll factor out the power of e that gets us the farthest from 0 as $x\to-\infty$. In this case, that's $e^{7x}$. $$\begin{align} \lim_{x\to-\infty}(2e^{6x}-e^{-7x}-10e^{4x}) &= \lim_{x\to-\infty}(e^{-7x}(2e^{13x}-1-10e^{11x}))\\ \end{align}$$ Here, the e's with positive exponents will go to 0 as $x\to-\infty$, and the $e^{-7x}$ goes to infinity, so we have $$\infty(2\cdot 0 - 1 - 10\cdot 0) = -\infty$$ for our final answer.
2. This time we'll factor out the $e^{6x}$, since that will be the largest as $x\to\infty$. $$\begin{align} \lim_{x\to-\infty}(2e^{6x}-e^{-7x}-10e^{4x}) &= \lim_{x\to-\infty}(e^{6x}(2-e^{-13x}-10e^{-2x})) = \infty(2) = \infty \end{align}$$



$f(x)=3e^{-x}-8e^{-5x}-e^{10x}$
Click for solution
1. To see what's happening as $x\to-\infty$, we can factor out the biggest power of e (in that direction), which is $e^{-5x}$: $$\begin{align} \lim_{x\to-\infty}(3e^{-x}-8e^{-5x}-e^{10x}) &= \lim_{x\to-\infty}(e^{-5x}(3e^{4x}-8-e^{15x}))\\ &= \infty(3\cdot 0 - 8 - 0 ) \\ &=-\infty\\ \end{align}$$
2. $$\begin{align} \lim_{x\to\infty}(3e^{-x}-8e^{-5x}-e^{10x}) &= \lim_{x\to\infty}(e^{10x}(3e^{-11x}-8e^{-15x}-1))\\ &= \infty(3\cdot 0 - 8\cdot 0 - 1 ) \\ &=-\infty\\ \end{align}$$



$f(x) = \frac{e^{-3x}-2e^{8x}}{9e^{8x}-7e^{-3x}}$
Click for solution
1. The exponential with the negative exponent is the only term in the denominator going to infinity for this limit and so we’ll need to factor the exponential with the negative exponent in the denominator from both the numerator and denominator to evaluate this limit. $$\begin{align} \lim_{x\to-\infty}\frac{e^{-3x}-2e^{8x}}{9e^{8x}-7e^{-3x}} &= \lim_{x\to-\infty}\frac{e^{-3x}(1-2e^{11x})}{e^{-3x}(9e^{11x}-7)}\\ &= \lim_{x\to-\infty}\frac{1-2e^{11x}}{9e^{11x}-7}\\ &=\frac{1-2\cdot0}{9\cdot0 -7}\\ &=-\frac17\\ \end{align}$$
2. The exponential with the positive exponent is the only term in the denominator going to infinity for this limit and so we’ll need to factor the exponential with the positive exponent in the denominator from both the numerator and denominator to evaluate this limit. $$\begin{align} \lim_{x\to\infty}\frac{e^{-3x}-2e^{8x}}{9e^{8x}-7e^{-3x}} &= \lim_{x\to\infty}\frac{e^{8x}((e^{-11x}-2)}{e^{8x}(9-7e^{-11x})}\\ &=\lim_{x\to\infty}\frac{e^{-11x}-2}{9-7e^{-11x}}\\ &=\frac{0-2}{9-7\cdot 0}\\ &=-\frac29\\ \end{align}$$



$f(x) = \frac{e^{-7x}-2e^{3x}-e^x}{e^{-x}+16e^{10x}+2e^{-4x}}$
Click for solution
1. The exponentials with the negative exponents are the only terms in the denominator going to infinity for this limit and so we’ll need to factor the exponential with the most negative exponent in the denominator (because it will be going to infinity fastest) from both the numerator and denominator to evaluate this limit. $$\begin{align} \lim_{x\to-\infty}\frac{e^{-7x}-2e^{3x}-e^x}{e^{-x}+16e^{10x}+2e^{-4x}} &= \lim_{x\to-\infty}\frac{e^{-4x}(e^{-3x}-2e^{7x}-e^{5x})}{e^{-4x}(e^{3x}+16e^{14x}+2)}\\ &= \lim_{x\to-\infty}\frac{e^{-3x}-2e^{7x}-e^{5x}}{e^{3x}+16e^{14x}+2}\\ &= \frac{\infty-0-0}{0+0+2}\\ &=\infty\\ \end{align}$$
2. The exponentials with the positive exponents are the only terms in the denominator going to infinity for this limit and so we’ll need to factor the exponential with the most positive exponent in the denominator (because it will be going to infinity fastest) from both the numerator and denominator to evaluate this limit. $$\begin{align} \lim_{x\to\infty}\frac{e^{-7x}-2e^{3x}-e^x}{e^{-x}+16e^{10x}+2e^{-4x}} &= \lim_{x\to\infty}\frac{e^{10x}(e^{-17x}-2e^{-7x}-e^{-9x})}{e^{10x}(e^{-11x}+16+2e^{-14x})}\\ &= \lim_{x\to\infty}\frac{e^{-17x}-2e^{-7x}-e^{-9x}}{e^{-11x}+16+2e^{-14x}}\\ &= \frac{0-0-0}{0+16+0}\\ &=0\\ \end{align}$$




For the following problems, evaluate the given limit.

$\lim\limits_{t\to-\infty}\ln(4-9t-t^3)$

Click for solution

Let $u = 4 - 9t - t^3$. Since $\lim\limits_{t\to-\infty}(4-9t-t^3) = \infty$, then $u\to \infty$ as $t \to -\infty$. Then our limit becomes $$\lim_{t\to-\infty}\ln(4-9t-t^3) = \lim_{u\to\infty}\ln(u) = \infty.$$





$\lim\limits_{z\to-\infty}\ln\left(\frac{3z^4-8}{2+z^2}\right)$

Click for solution

Let $t = \frac{3z^4-8}{2+z^2}$. We want to find what happens to $t$ as $z \to -\infty$: $$\begin{align} \lim_{z\to-\infty}\frac{3z^4-8}{2+z^2} &= \lim_{z\to-\infty}\frac{z^2(3z^2-\frac{8}{z^2})}{z^2(\frac{2}{z^2}+1)}\\ &=\lim_{z\to-\infty}\frac{3z^2-\frac{8}{z^2}}{\frac{2}{z^2}+1}\\ &=\frac{\infty - 0}{0+1}\\ &=\infty\\ \end{align}$$ Therefore, $t\to\infty$. So our original limit can be evaluated as such: $$$\lim\limits_{z\to-\infty}\ln\left(\frac{3z^4-8}{2+z^2}\right) = \lim_{t\to \infty}\ln(t) = \infty.$$





$\lim\limits_{x\to\infty}\ln\left(\frac{11+8x}{x^3+7x}\right)$

Click for solution

Let $t = \frac{11+8x}{x^3+7x}$. $$\begin{align} \lim_{x\to\infty}\frac{11+8x}{x^3+7x} &= \lim_{x\to\infty}\frac{x^3(\frac{11}{x^3}+\frac{8}{x^2})}{x^3(1+\frac{7}{x^2})}\\ &=\lim_{x\to\infty}\frac{\frac{11}{x^3}+\frac{8}{x^2}}{1+\frac{7}{x^2}}\\ &= \frac{0+ 0}{1+0}\\ &=0\\ \end{align}$$ Also, note that because we are evaluating the limit $x\to\infty$ it is safe to assume that $x\gt 0$ and so we can further say that as $x\to\infty$, $t\to0^+$. Our original limit then becomes $$\lim\limits_{x\to\infty}\ln\left(\frac{11+8x}{x^3+7x}\right) = \lim_{t\to0}\ln(t) = -\infty.$$





$\lim\limits_{x\to-\infty}\tan^{-1}(7-x+3x^5)$

Click for solution

We'll let $t = 7-x+3x^5$. We see that $$\begin{align} \lim\limits_{x\to-\infty}(7-x+3x^5) &= \lim_{x\to-\infty}(x^5(\frac{7}{x^5}-\frac{1}{x^4} + 3))\\ &= -\infty(0-0+3)\\ &=-\infty\\ \end{align}$$ so as $x\to-\infty$, we also have $t\to-\infty$. Therefore, our limit is $$\lim\limits_{x\to-\infty}\tan^{-1}(7-x+3x^5) = \lim_{t\to-\infty}\tan^{-1}(t) = -\frac{\pi}{2}.$$





$\lim\limits_{t\to\infty}\tan^{-1}\left(\frac{4+7t}{2-t}\right)$

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Let $u = \frac{4+7t}{2-t}$. We'll figure out what's happening to that input first. $$\begin{align} \lim_{t\to\infty}\frac{4+7t}{2-t} &= \lim_{t\to\infty}\frac{t(\frac{4}{t}+7)}{t(\frac{2}{t}-1)}\\ &=\lim_{t\to\infty}\frac{\frac{4}{t}+7}{\frac{2}{t}-1}\\ &=\frac{0+7}{0-1}\\ &= -7\\ \end{align}$$ The answer, then, is $$\lim\limits_{t\to\infty}\tan^{-1}\left(\frac{4+7t}{2-t}\right) = \lim_{u\to -7}\tan^{-1}(u) = \tan^{-1}(-7).$$ The inverse tangent function is continuous on its domain, and -7 is in that domain, so we can use direct substitution to evaluate this limit! Do not get so used the “special case” limits that we tend to usually do in the problems at the end of a section that you decide that you must have done something wrong when you run across a problem that doesn’t fall in the “special case” category.





$\lim\limits_{w\to\infty}\tan^{-1}\left(\frac{3w^2-9w^4}{4w-w^3}\right)$

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Let $t = \frac{3w^2-9w^4}{4w-w^3}$. $$\begin{align} \lim_{w\to\infty}\frac{3w^2-9w^4}{4w-w^3} &= \lim_{w\to\infty}\frac{w^3(\frac{3}{w}-9w)}{w^3(\frac{4}{w^2}-1)}\\ &=\lim_{w\to\infty}\frac{\frac{3}{w}-9w}{\frac{4}{w^2}-1}\\ &= \frac{0-\infty}{0-1}\\ &=\infty\\ \end{align}$$ The limit we're interested in then becomes $$\lim\limits_{w\to\infty}\tan^{-1}\left(\frac{3w^2-9w^4}{4w-w^3}\right) = \lim_{t\to\infty}\tan^{-1}(t) = \frac{\pi}{2}.$$

Assignment Problems

For the following problems, evaluate (a)$\lim\limits_{x\to-\infty}f(x)$ and (b)$\lim\limits_{x\to\infty}f(x)$.

$f(x) = e^{(x^4+8x)}$
$f(x) = e^{(2x+4x^2+2x^5)}$
$f(x) = e^{\left(\frac{3-x^3}{x+x^2}\right)}$
$f(x) = e^{\left(\frac{5-9x}{7+3x}\right)}$
$f(x) = e^{\left(\frac{5+2x^6}{x-8x^4}\right)}$
$f(x) = e^x+12e^{-3x}-2e^{-10x}$
$f(x) = 9e^{2x}-7e^{-14x}-e^x$
$f(x) = 20e^{-8x}-e^{5x}+3e^{2x}-e^{-7x}$
$f(x) = \frac{6e^{4x}+e^{-15x}}{11e^{4x}+6e^{-15x}}$
$f(x) = \frac{e^{3x}+9e^{-x}-4e^{10x}}{2e^{7x}-e^{-x}}$
$f(x) = \frac{3e^{-14x}-e^{18x}}{e^{-x}-2e^{20x}-e^{-9x}}$



For the following problems, evaluate the given limit.

$\lim\limits_{x\to\infty}\ln(5x^2+12x-6)$
$\lim\limits_{y\to-\infty}\ln(5-7y^5)$
$\lim\limits_{x\to\infty}\ln\left(\frac{3+x}{1+5x^3}\right)$
$\lim\limits_{t\to-\infty}\ln\left(\frac{2t-5t^3}{4+3t^2}\right)$
$\lim\limits_{z\to\infty}\left(\frac{10z+8z^2}{z^2-1}\right)$
$\lim\limits_{x\to-\infty}\tan^{-1}(7+4x-x^3)$
$\lim\limits_{w\to\infty}\tan^{-1}(4w^2-w^6)$
$\lim\limits_{t\to\infty}\tan^{-1}\left(\frac{4t^3+t^2}{1+3t}\right)$
$\lim\limits_{z\to-\infty}\tan^{-1}\left(\frac{z^4+4}{3z^2+5z^3}\right)$