C.7 Limits of Exponential and Logarithmic Functions

On the preceding screens we looked at the behavior of polynomials and rational functions as 𝑥 → ±∞, 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 →∞ and 𝑥 → −∞ separately, starting with the former.

Exponentials as 𝑥 →∞

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

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

Exponentials as 𝑥 → −∞

When thinking about the behavior of exponential functions, one helpful strategy is to make the substitution 𝑡 = −𝑥 and then consider the limit lim𝑡→∞𝑓(𝑡) instead. The following example illustrates.

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

Exponentials dominate power functions

Although we can't prove it — yet — we hope you'll trust us when we say that any exponential 𝑎𝑥 (with 𝑎 >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.

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.

Summary: Any exponential 𝑎𝑥 (with 𝑎 >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 𝑥𝑟 (for 𝑟 ≥0) dominates over the logarithmic function ln⁡𝑥.

While the exploration above shows that as 𝑥 →∞ 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.

Summary: Any power function 𝑥𝑟 (for 𝑟 >0) dominates over any logarithmic function.

Practice Problems: Limits at Infinity of Exponential and Logarithmic Functions

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

Practice Problem 1

Find lim𝑥→∞𝑥+ln⁡𝑥5𝑒𝑥.

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lim𝑥→∞𝑥+ln⁡𝑥5𝑒𝑥=0✓ because the 𝑒𝑥 in the denominator dominates over both of the terms in the numerator.

Practice Problem 2

Find lim𝑥→∞2𝑥+3𝑥3𝑥.

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In the numerator, the 3𝑥 dominates over the 2𝑥, so lim𝑥→∞(2𝑥+3𝑥)=lim𝑥→∞3𝑥. Hence lim𝑥→∞2𝑥+3𝑥3𝑥=lim𝑥→∞3𝑥3𝑥=1✓ As the graph shows, the horizontal line 𝑦 =1 is thus a horizontal asymptote as 𝑥 →∞.

Practice Problem 3

Find lim𝑥→∞9𝑥2−6𝑥+1𝑥ln⁡𝑥+log⁡𝑥.

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lim𝑥→∞9𝑥2−6𝑥+1𝑥ln⁡𝑥+log⁡𝑥=lim𝑥→∞9𝑥2ln⁡𝑥+log⁡𝑥=∞✓ because the 9𝑥2 in the numerator dominates over both log-functions in the denominator.

Practice Problem 4

Find lim𝑥→∞𝑥𝑒−𝑥 and lim𝑥→−∞𝑥𝑒−𝑥.

Open for hint.

For 𝑥 →∞ :The function might be easier to think about as 𝑥𝑒−𝑥 =𝑥𝑒𝑥.
For 𝑥 → −∞: Consider making the substitution 𝑥 = −𝑡. collapse

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Let's think first about 𝑥 →∞: We know that the exponential 𝑒−𝑥 dominates over the power function 𝑥. Hence lim𝑥→∞𝑥𝑒−𝑥=lim𝑥→∞𝑒−𝑥=0✓ A related way to view the question is as suggested by the hint: lim𝑥→∞𝑥𝑒−𝑥=lim𝑥→∞𝑥𝑒𝑥=0✓ since the dominating 𝑒𝑥 is in the denominator.

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Now for 𝑥 → −∞: lim𝑥→−∞𝑥𝑒−𝑥=(lim𝑥→−∞𝑥)(lim𝑥→−∞𝑒−𝑥)=(−∞)(∞)=−∞✓ where in the last line we use the fact that as 𝑥 → −∞, 𝑒−𝑥 takes on larger and Larger negative input values and returns larger and LARGER positive output values (because 𝑒(anything) always returns a positive value). For instance, 𝑓( −1000) =( −1000)𝑒−(−1000) = −1000𝑒1000, and 𝑓( −10000) =( −10000)𝑒−(−10000) = −10000𝑒10000, and so forth.

We can also use out substitution 𝑥 = −𝑡 to examine the behavior 𝑥 → −∞ instead as 𝑡 → = +∞ : lim𝑥→−∞𝑥𝑒−𝑥=lim𝑡→∞(−𝑡)𝑒𝑡=(−∞)(∞)=−∞✓ In words: as 𝑥 → −∞, 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 −∞.

The Upshot

1. We can summarize our discussion of dominance from the preceding several screens as: 𝑎𝑥 dominates 𝑥𝑟 dominates ln⁡𝑥 In words: exponentials dominate power functions (and polynomials) dominate logarithms.

Content Note

Some courses, but not all, require that you know how to find the limit as 𝑥 → ±∞ 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 entire 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 ±∞. You now have many tools to reason about a function's behavior as 𝑥 → −∞ and 𝑥 →∞!

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 → ±∞? Please join the community over on the Forum so we can all put our heads together!