C.5 Limits of Polynomials

On this screen we're going to see how to reason quickly about the limit of polynomials as x goes to positive or negative infinity, relying on a concept known as "dominance" as applied to polynomials. We'll also show you how to prove such results more formally. For both reasoning paths we'll make direct use of the conclusions from the preceding two screens (limits as some limits as 𝑥 →∞ and some limits as 𝑥 → −∞).

More specifically, on the preceding screens we developed the skills we need to determine the limits of a monomial, like lim𝑥→∞𝑥3=∞ and lim𝑥→∞(−35𝑥2)=−35lim𝑥→∞𝑥2=−∞. What then is the limit as 𝑥 →∞ of a polynomial like 𝑓(𝑥) =𝑥3 −35𝑥2 +16𝑥, where the limit of the first and third terms go to +∞ but the limit of the second term goes to −∞? Is the limit of the whole function +∞, or −∞, or 0, or something else?

Let's explore the behavior of this function to illustrate the general approach to finding such limits.

Explore 1: lim𝑥→∞(𝑥3−35𝑥2+16𝑥)

Consider the function 𝑓(𝑥) =𝑥3 −35𝑥2 +16𝑥.

What do you think its limit is as 𝑥 →∞?
Even if you're not sure of the answer, what features of the function are you drawn to think about in considering its behavior as x grows and Grows and GROWS larger and Larger still?

Informal, quick reasoning
One thought you might have had is that behavior of the term with the largest power, 𝑥3, dominates, or "wins," over the other terms.

Such an insight is correct, as you can see using the interactive graph below: The solid red curve shows 𝑓(𝑥), while the dashed-line blue curve shows only 𝑔(𝑥) =𝑥3 for comparison purposes. Initially the two curves are quite different. But as you zoom progressively outward — as you can do easily using the "Zoom Out" buttons beneath the graph — you can see that the two curves become more and more indistinguishable. In effect, the smaller-power terms become less and less important as x grows without bound.

For example, at Zoom Level 4, the two curves look like they overlap entirely for large enough x. Let's consider the two functions' outputs at 𝑥 =10 000 to illustrate numerically: 𝑓(𝑥)=𝑥3−35𝑥2+16𝑥𝑓(10000)=(10000)3−35(10000)2+16(10000)=1.0×1012−3.5×109+1.6×105

Looking term-by-term, we see that at 𝑥 =10 000 the 𝑥3 term is of order 1012, while the second, next-largest term is only of order 109, so an effect approximately 1 000-times as small. The third term, 16𝑥, is much smaller still: of order 105 it is not a small number, but is rather insignificant compared to the first term's size of 1012.

Furthermore, remember that 𝑥 →∞ means x can be as large as we'd like. So if we consider an even larger input value, say 𝑥 =1020, then we have 𝑓(𝑥)=𝑥3−35𝑥2+16𝑥𝑓(1020)=(1020)3−35(1020)2+16(1020)=1.0×1060−3.5×1041+1.6×1021 Here the second term is approximately 1019-times as small as the first term, and the third terms is another 1020-times as small.

This is all to say that as x grows larger, the term with the largest power (𝑥3) dominates over the other terms with smaller powers (−35𝑥2 and 16𝑥). Hence when thinking about the limit as 𝑥 →∞, we can focus only on that largest term that dominates, since its limit is the same as the limit as the original function:

𝑥3−35𝑥2+16𝑥as 𝑥 grows large←←←←←←←←←←←←←←←←←→𝑥3 Then simply recall that lim𝑥→∞𝑥3=∞, which lets you quickly reason that lim𝑥→∞(𝑥3−35𝑥2+16𝑥)=lim𝑥→∞𝑥3=∞✓

This informal reasoning will allow you to determine the limit quickly and is often the only approach you need: you read the question, look at the polynomial, decide what dominates, and immediately know the answer. You'll of course be able to practice this below.

Formal development of the limit

The informal reasoning above is clearly not a rigorous determination of the limit. We can do that easily enough as well, though, using a simple tactic: First factor out the largest-power term, in this case the 𝑥3: 𝑥3−35𝑥2+16𝑥=𝑥3(1−35𝑥+16𝑥2)

Then recall that lim𝑥→∞1𝑥=0 and lim𝑥→∞1𝑥2=0: lim𝑥→∞(𝑥3−35𝑥2+16𝑥)=lim𝑥→∞[𝑥3(1−35𝑥+16𝑥2)]=[lim𝑥→∞𝑥3][lim𝑥→∞1−35lim𝑥→∞1𝑥+16lim𝑥→∞1𝑥2]=[lim𝑥→∞𝑥3][lim𝑥→∞1−35lim𝑥→∞1𝑥0+16lim𝑥→∞1𝑥20]=[lim𝑥→∞𝑥3][1][Recall lim𝑥→∞𝑥3=∞]=∞✓

And that's our proof that the limit of this polynomial is ∞, confirming our informal reasoning.

Dominance: focus only on the term with the largest power

The key point of the Exploration above is that, when considering lim𝑥→∞, we can focus only on the term in the polynomial with the largest power. This approach is known as dominance: the largest-power term dominates, or "wins," over all of the other terms as 𝑥 →∞.

Graph illustrating the concept of dominance: the original function f(x) = x^3 -35x^2 + 16x, and the function g(x) = x^3, become essentially identical for large x.

For instance, for the function above 𝑓(𝑥) =𝑥3 −35𝑥2 +16𝑥, we need only consider the 𝑥3 term for our reasoning.

Let's summarize how we use the concept of dominance both informally and formally:

Dominance and Polynomials, lim𝑥→∞𝑓(𝑥)

For quick informal reasoning, use dominance and just look at the largest term: lim𝑥→∞(𝐴𝑥𝑁+ (smaller terms))=lim𝑥→∞𝐴𝑥𝑁 and then use what you know about that limit from the preceding screens.

For formal development of the limit: Factor out the largest-power term that dominates, and then use what you know about the resulting limits for lim𝑥→∞𝑥𝑛 and lim𝑥→∞1𝑥𝑛.

You'll be able to practice these approaches below. Before then, let's extend our ideas to apply to 𝑥 → −∞, using the same function as above. Since the reasoning regarding dominance is largely the same, we'll proceed more quickly.

Explore 2: lim𝑥→−∞(𝑥3−35𝑥2+16𝑥)

What is lim𝑥→−∞(𝑥3−35𝑥2+16𝑥)?

You can again use the interactive Desmos calculator below, and the "Zoom Out" buttons, to explore your initial thoughts.

Informal, quick reasoning
As above, when thinking about the limit we can focus only on the largest term that dominates: 𝑥3−352𝑥2+16𝑥as 𝑥 grows large←←←←←←←←←←←←←←←←←→𝑥3 and then recall that lim𝑥→−∞𝑥3=−∞, and so lim𝑥→−∞(𝑥3−35𝑥2+16𝑥)=lim𝑥→−∞𝑥3=−∞✓

Formal development of the limit
Also as above, we first factor out the largest-power term that dominates, and then recall that lim𝑥→−∞1𝑥=0 and lim𝑥→−∞1𝑥2=0: lim𝑥→−∞(𝑥3−35𝑥2+16𝑥)=lim𝑥→−∞[𝑥3(1−35𝑥+16𝑥2)]=[lim𝑥→−∞𝑥3][lim𝑥→−∞1−35lim𝑥→−∞1𝑥+16lim𝑥→−∞1𝑥2]=[lim𝑥→−∞𝑥3][lim𝑥→−∞1−35lim𝑥→−∞1𝑥0+16lim𝑥→−∞1𝑥20]=[lim𝑥→−∞𝑥3][1][Recall lim𝑥→−∞𝑥3=−∞]=−∞✓

Exactly as above, we can summarize the reasoning for both informal and formal development of the limit:

Dominance and Polynomials, lim𝑥→−∞𝑓(𝑥)

For quick informal reasoning: lim𝑥→−∞(𝐴𝑥𝑁+ (smaller terms))=lim𝑥→−∞𝐴𝑥𝑁

For formal development of the limit, factor out the largest-power term and then use what you know about the resulting limits for lim𝑥→−∞𝑥𝑛 and lim𝑥→−∞1𝑥𝑛.

Of course there are other types of functions in the world besides polynomials; for instance, rational functions (as you know) are fractions of polynomials. We'll see how to extend the ideas of dominance to determine the limit as 𝑥 → ±∞ of such rational functions on the next screen!

For now, practice using the ideas from this screen in the following few problems. We recommend (1) quickly deciding the answer using informal reasoning, and then (2) put pencil-to-paper and to practice finding the limit more formally, which you'll probably need to do on an exam. You'll discover the solution only takes 2-3 lines, and once you've done a few of these you'll have the technique down for yourself.

Practice Problem 1
(a)
lim𝑥→∞(3𝑥3+947𝑥2−√𝑥)
(b)
lim𝑥→−∞(3𝑥3+947𝑥2−√𝑥)
Practice Problem 2
Find lim𝑥→∞(𝑥−𝑥2).

The Upshot

  1. When considering the limit on a polynomial as 𝑥 →∞ or 𝑥 → −∞, focus only on the term with the largest power — a concept known as "dominance," since that largest power dominates over all of the others for large x.
  2. With that focus in place, you can immediately deduce the limit of any polynomial, after recalling (or picturing in your head) the behavior of that largest term: lim𝑥→∞(𝐴𝑥𝑁+ (smaller terms))=lim𝑥→∞𝐴𝑥𝑁 lim𝑥→−∞(𝐴𝑥𝑁+ (smaller terms))=lim𝑥→−∞𝐴𝑥𝑁
  3. To develop the required limit more formally, factor out the largest-power term and then proceed.

Please join the discussion over on the Forum for any questions about dominance, or any other concepts involving lim𝑥→±∞. We're waiting for you there!