C.7 Limits of Exponential and Logarithmic Functions

Calculus Home Limits & Continuity C. Limits at Infinity C.7 Limits of Exponential and Logarithmic Functions

On the preceding screens we looked at the behavior of polynomials and rational functions as $x \to \pm \infty,$ and saw that the largest power dominates in each case. Let’s now look at the limits of exponential functions and logarithmic functions into this mix, and see what functions they dominate and what functions dominate them.

Exponential Functions

We’ll consider the limits as $x \to \infty$ and $x \to -\infty$ separately, starting with the former.

Exponentials as $x \to \infty$
It won’t surprise you that if we compare two exponential functions, like $f(x) = 2^x$ and $g(x) = 3^x,$ the one with the larger base grows faster and hence dominates as $x \to \infty.$ You can see this for yourself using the interactive graph in the following Exploration.

Expoloration 1: Comparing exponential functions with different bases

You can use the sliders beneath the graph to change the base of each function to help visualize how the exponential function with the larger base grows more rapidly as $x \to \infty.$

Two exponential growth curves to illustrate that
the one with the larger base grows faster as $x \to \infty$

Currently $f(x)$ = $2^x$
(solid red curve)

Currently $g(x)$ = $3^x$
(dashed blue curve)

The simple insight that the exponential with the largest base dominates lets us quickly answer questions like the following.

Example 1. $\displaystyle{\lim_{x \to \infty}\dfrac{2^x + 1}{3^x +7}}$ & $\displaystyle{\lim_{x \to \infty}\dfrac{3^x + 1}{2^x +7}}$

Find the requested limits.

Find $\displaystyle{\lim_{x \to \infty}\dfrac{2^x + 1}{3^x +7}}.$

Find $\displaystyle{\lim_{x \to \infty}\dfrac{3^x + 1}{2^x +7}}.$

Solution.
(a)
\[\lim_{x \to \infty}\frac{2^x + 1}{3^x +7} = \lim_{x \to \infty}\frac{2^x}{3^x} = 0 \quad \cmark\] since the larger exponential, $3^x$, is in the denominator. For the last piece of reasoning there you could also realize that
\[\lim_{x \to \infty}\frac{2^x}{3^x} = \lim_{x \to \infty} \left(\frac{2}{3}\right)^x = 0 \quad \cmark\] since $\left(\dfrac{2}{3} \right)^x \to 0$ as $x \to \infty$ as the fraction $\dfrac{2}{3}$ multiplies itself again and again.
(b)
\[ \lim_{x \to \infty}\frac{3^x + 1}{2^x +7} = \lim_{x \to \infty}\frac{3^x}{2^x} = \infty \quad \cmark\] since the larger exponential, $3^x$, is in the numerator. You could also reason that
\[\lim_{x \to \infty}\frac{3^x}{2^x} = \lim_{x \to \infty}\left( \frac{3}{2} \right)^x = \infty \quad \cmark\] since $\left( \dfrac{3}{2} \right)^x$ grows and Grows and GROWS as $x \to \infty$ as the fraction $\dfrac{3}{2}$ multiplies itself again and again.

The graph below shows how when the exponential with the larger base is in the denominator the function goes to 0 as $x \to \infty,$ whereas when the exponential with the larger base in in the numerator the function goes to $\infty.$

Graphs of $f(x) =\dfrac{2^x + 1}{3^x +7}$ and $g(x) = \dfrac{3^x + 1}{2^x +7}$

Exponentials as $x \to -\infty$
When thinking about the behavior of exponential functions, one helpful strategy is to make the substitution $t = -x$ and then consider the limit $\displaystyle{\lim_{t \to \infty}f(t)}$ instead. The following example illustrates.

Example 2. $\displaystyle{\lim_{x \to \infty}\dfrac{4^x}{5^x}}$ and $\displaystyle{\lim_{x \to -\infty}\dfrac{4^x}{5^x}}$

Find the requested limits.

Find $\displaystyle{\lim_{x \to \infty}\dfrac{4^x}{5^x}}$.

Find $\displaystyle{\lim_{x \to -\infty}\dfrac{4^x}{5^x}}$.

Solution.
(a) \[ \lim_{x \to \infty}\dfrac{4^x}{5^x} = 0 \quad \cmark \] since the dominating exponential, $5^x$, is in the denominator.

Another way to view this function is
\[ \dfrac{4^x}{5^x} = \left(\dfrac{4}{5} \right)^x\] in which case we can imagine as we multiply $\dfrac{4}{5}$ by itself again and again and again and (so on) as $x \to \infty,$ the output becomes closer and closer to 0:
\[ \lim_{x \to \infty}\dfrac{4^x}{5^x} = \lim_{x \to \infty}\left(\dfrac{4}{5} \right)^x = 0 \quad \cmark \]

(b) We’re going to think about this question of $x \to -\infty$ in several related ways.
First, let’s use the suggestion to make the substitution $x = -t,$ which means we’re now looking at the limit as $t \to +\infty$:
\begin{align*}
\lim_{x \to -\infty}\dfrac{4^x}{5^x} &= \lim_{t \to \infty}\dfrac{4^{-t}}{5^{-t}} \\[8px] &= \lim_{t \to \infty}\dfrac{5^{t}}{4^{t}} \\[8px] &= \infty \quad \cmark
\end{align*}
since the dominating exponential $5^t$ is now in the numerator and we’re thinking about the limit as $t \to \infty$ (which we just find easier here).

As a related way to view this result, we have
\[ \lim_{x \to -\infty}\dfrac{4^x}{5^x} = \lim_{t \to \infty}\dfrac{4^{-t}}{5^{-t}} = \lim_{t \to \infty}\left(\frac{5}{4} \right)^t = \infty \quad \cmark\] since we’re now multiplying the fraction $\dfrac{5}{4}$ by itself again and again and again as $t \to \infty.$

We can also develop this result by reasoning with the original function $\dfrac{4^x}{5^x}$ instead, if you’d rather:
As $x \to -\infty$, the behavior of the exponential $5^x$ still dominates, but now in that it goes to zero faster than $4^x$ goes to zero. Hence the fraction $\dfrac{4^x}{5^x}$ blows up as $x \to -\infty.$

Graph of $f(x) = \dfrac{4^x}{5^x}$ versus x

Summary: As $x \to \infty,$ the exponential with a larger base dominates an exponential with a smaller base.
If you’re looking for the limit as $x \to -\infty$ of a function with exponentials, it may be helpful to make the substitution $t = -x$ and think about $t \to +\infty$ instead.

Exponentials dominate power functions

Although we can’t prove it — yet — we hope you’ll trust us when we say that any exponential $a^x$ (with $a \gt 1$) dominates over any power function or polynomial.

The following Exploration isn’t meant as a proof, but rather a chance to see how this works in practice.

Exploration 2: Comparing $f(x) = 2^x$ and $g(x) = x^5$

The interactive graph below shows the two functions $f(x) = 2^x$ and $g(x) = x^5.$ Offhand, you might think that the $x^5$ function dominates: after all, at $x = 2$
\[ g(2) = 2^5 = 32 \quad \text{while}\quad f(2) = 2^2 = 4\] as shown initially in the graph below.

Graphs of $f(x) = 2^x$ and $g(x) = x^5$

But we’re telling you: as $x \to \infty,$ the exponential function dominates over the power function. To see that this is the case here, hit the “Zoom Out” button and you’ll see that as $x \to \infty,$ the exponential function $2^x$ “catches up to” and then overtakes the power function $x^5.$ Notice that this happens around $x = 22.5,$ which isn’t even a very large value of x (although the y-values are large). Furthermore, since the exponential is growing at the faster rate, it will dominate forever after, as the graph suggests.

With the knowledge that exponentials (with base greater than 1) grow faster than any power function, we can easily answer questions like those in the following example.

Example 3: $\displaystyle{\lim_{x \to \infty}\dfrac{x^3 + 7x^2 +1}{1.1^x}}$ and $\displaystyle{\lim_{x \to \infty}\dfrac{1.1^x}{x^3 + 7x^2 +1}}$

Find the requested limits.

Find $\displaystyle{\lim_{x \to \infty}\dfrac{x^3 + 7x^2 +1}{1.1^x}}.$

Find $\displaystyle{\lim_{x \to \infty}\dfrac{1.1^x}{x^3 + 7x^2 +1}}.$

Solution.
(a)
\[\lim_{x \to \infty}\dfrac{x^3 + 7x^2 +1}{1.1^x} = \lim_{x \to \infty}\dfrac{x^3}{1.1^x} = 0 \quad \cmark\] since the exponential in the denominator dominates over the polynomial in the numerator.
(b)
\[\lim_{x \to \infty}\dfrac{1.1^x}{x^3 + 7x^2 +1} = \lim_{x \to \infty}\dfrac{1.1^x}{x^3} = \infty \quad \cmark\] since the exponential in the numerator dominates over the polynomial in the denominator.

The interactive graph below shows the two functions. Notice that the initial graphing window — similar to what you would see if you just opened a Desmos calculator and plotted these functions — is quite misleading for showing the limit as $x \to \infty.$ For a more-correct image, click the “Zoom out” button beneath the graph.

Graphs of $f(x) = \dfrac{x^3 + 7x^2 +1}{1.1^x}$ and $g(x) = \dfrac{1.1^x}{x^3 + 7x^2 +1}$

Summary: Any exponential $a^x$ (with $a \gt 1$) dominates over any power function or polynomial.

Power functions dominate logarithms

Finally, although we again can’t prove it yet, you can see in the Exploration below that any power function $x^r$ (for $r \ge 0$) dominates over the logarithmic function $\ln x.$

Exploration 3: Comparing $f(x) = x^r$ and $g(x) = \ln x$

The graph below plots $f(x) = x^r$ (solid green curve) and $g(x) = \ln x$ (dashed blue curve).

You can use the slider beneath the graph to change the value of the power-function exponent r, and easily see that for $r \ge 1$ the power function grows much more rapidly than $\ln x.$

Graphs of $f(x) = x^r$ and $g(x) = \ln x$

Currently $f(x)$ = $x^{1}$

After you’ve seen that for $r \ge 1$ the power function dominates, please set $r = 0.1.$
While it might appear that for a small power like $x^{0.1}$ the $\ln x$ dominates, even here the power function eventually overtakes the $\ln x$: you can use the “Zoom way out” button below to see where this occurs. Past this point (approximately $x \approx 3.4 \times 10^{15}$) the power function is always larger than the logarithmic function.

For use when $r = 0.1$ and so $f(x) = x^{0.1}:$

By the way, with $r = 0.1$ and $f(x) = x^{0.1}$ you can see just how slowly both of these functions grow: for input values around $x = 5 \times 10^{15},$ the output values are around . . . 35!

Let’s not lose sight of the key takeaway here: any power function $x^r$ (where $r \gt 0$) dominates over the logarithmic function.

While the exploration above shows that as $x \to \infty$ each of the power functions available dominates over the natural logarithm, $\ln x,$ the same conclusion holds for logarithms of any base. That is, any power function dominates over any logarithmic function.

With this knowledge about dominance over power functions over logs, we can easily answer questions like those in the following example.

Example 4: $\displaystyle{\lim_{x \to \infty}\dfrac{\log_2 x}{x+1}}$ and $\displaystyle{\lim_{x \to \infty}\dfrac{x+1}{\log_2 x}}$

Find the requested limits.

Find $\displaystyle{\lim_{x \to \infty}\dfrac{\log_2 x}{x+1}}.$

Find $\displaystyle{\lim_{x \to \infty}\dfrac{x+1}{\log_2 x}}.$

Solution.
(a)
\[\lim_{x \to \infty}\dfrac{\log_2 x}{x+1} = \lim_{x \to \infty}\dfrac{\log_2 x}{x} = 0 \quad \cmark\] because the dominating power function, $x^1,$ is in the denominator of the function.
(b)
\[\lim_{x \to \infty}\dfrac{x+1}{\log_2 x} = \lim_{x \to \infty}\dfrac{x}{\log_2 x} = \infty \quad \cmark\] because the dominating power function is in the numerator of the function.

Graphs of $f(x) = \dfrac{\log_2 x}{x+1}$ and $g(x) = \dfrac{x+1}{\log_2 x}$

Summary: Any power function $x^r$ (for $r \gt 0$) dominates over any logarithmic function.

Practice Problems

As usual, let’s consider a few practice problems to help cement your new knowledge.

Practice Problem #1

Find $\displaystyle{\lim_{x \to \infty}\dfrac{x + \ln x}{5e^x}}.$

Show/Hide Solution

\[ \lim_{x \to \infty}\dfrac{x + \ln x}{5e^x} = 0 \quad \cmark\] because the $e^x$ in the denominator dominates over both of the terms in the numerator.

Graph of the function, showing that it quickly heads to y = 0 due to the 5e^x in the denominator

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Practice Problem #2

Find $\displaystyle{\lim_{x \to \infty}\dfrac{2^x + 3^x}{3^x}}.$

Show/Hide Solution

In the numerator, the $3^x$ dominates over the $2^x,$ so $\displaystyle{\lim_{x \to \infty}\left(2^x + 3^x \right) = \lim_{x \to \infty}3^x. }$ Hence \[\lim_{x \to \infty}\dfrac{2^x + 3^x}{3^x} = \lim_{x \to \infty}\dfrac{3^x}{3^x} = 1 \quad \cmark\] As the graph shows, the horizontal line $y=1$ is thus a horizontal asymptote as $x \to \infty.$

Graph showing the function goes quickly to the horizontal line y=1 as x increases to infinity

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Practice Problem #3

Find $\displaystyle{\lim_{x \to \infty}\frac{9x^2 - 6x + \dfrac{1}{x}}{\ln x + \log x}}.$

Show/Hide Solution

\[\lim_{x \to \infty}\frac{9x^2 – 6x + \dfrac{1}{x}}{\ln x + \log x} = \lim_{x \to \infty}\frac{9x^2 }{\ln x + \log x} = \infty \quad \cmark \] because the $9x^2$ in the numerator dominates over both log-functions in the denominator.

Graph of the function showing that as x goes to infinity the output grows without bound

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Practice Problem #4

Find $\displaystyle{\lim_{x \to \infty}xe^{-x}}$ and $\displaystyle{\lim_{x \to -\infty}xe^{-x}.}$

Open for hint.

For $x \to \infty: $The function might be easier to think about as $xe^{-x} = \dfrac{x}{e^x}.$
For $x \to -\infty$: Consider making the substitution $x = -t.$

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Show/Hide Solution

Let’s think first about $x \to \infty$: We know that the exponential $e^{-x}$ dominates over the power function $x.$ Hence \[\lim_{x \to \infty}xe^{-x} = \lim_{x \to \infty}e^{-x} = 0 \quad \cmark\] A related way to view the question is as suggested by the hint: \[\lim_{x \to \infty}xe^{-x} = \lim_{x \to \infty}\frac{x}{e^x} = 0 \quad \cmark\] since the dominating $e^x$ is in the denominator.

Now for $x \to -\infty$: \begin{align*} \lim_{x \to -\infty}xe^{-x} &= \left(\lim_{x \to -\infty}x\right) \left( \lim_{x \to -\infty}e^{-x}\right) \\[8px] &= (-\infty)(\infty) = -\infty \quad \cmark \end{align*} where in the last line we use the fact that as $x \to -\infty,$ $e^{-x}$ takes on larger and Larger negative input values and returns larger and LARGER positive output values (because $e^{\text{(anything)}}$ always returns a positive value). For instance, $f(-1000) = (-1000)e^{-(-1000)} = -1000e^{1000},$ and $f(-10000) = (-10000)e^{-(-10000)} = -10000e^{10000},$ and so forth.

We can also use out substitution $x = -t$ to examine the behavior $x \to -\infty$ instead as $t \to = +\infty:$ \begin{align*} \lim_{x \to -\infty}xe^{-x} &= \lim_{t \to \infty}(-t)e^{t} \\[8px] &= (-\infty)(\infty) = -\infty \quad \cmark \end{align*} In words: as $x \to -\infty,$ $x$ grows without bound in the negative direction, while $e^{-x}$ grows without bound with ever-increasing positive output values. Hence the product of the two is a negative number that you can make as large as you’d like, and so the limit is $-\infty.$

Graph of f(x) = xe^(-x), showing that as x goes to infinity the curve goes to zero (the x-axis), while as x goes to negative infinity the curve goes very quickly to very large and ever-increasing negative values.

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The Upshot

We can summarize our discussion of dominance from the preceding several screens as:
\[ a^x \text{ dominates } x^r \text{ dominates } \ln x \] In words: exponentials dominate power functions (and polynomials) dominate logarithms.

Note: Some courses, but not all, require that you know how to find the limit as $x \to \pm \infty$ of functions that have a square root in it. While not conceptually hard, there are some subtleties to computing these limits that require special attention, and so we’ve dedicated the next screen to this tricky topic. Please ask your instructor if this is something you need to be able to do, and if so continue to there. If not, please proceed to the next Section, as described below!

This concludes our exploration of limits at $\pm \infty.$ You now have many tools to reason about a function’s behavior as $x \to -\infty$ and $x \to \infty!$

In the next section, we’ll take up the important concept of “continuity” and “continuous functions,” along with the related “Intermediate Value Theorem.” We’ll see you there. 🙂

In the meantime, what questions, thoughts or comments do you have about limits as $x \to \pm \infty$? Please join the community over on the Forum so we can all put our heads together!

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