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 $x \to \infty$ and some limits as $x \to -\infty$).

More specifically, on the preceding screens we developed the skills we need to determine the limits of a monomial, like $\displaystyle{\lim_{x \to \infty} x^3 = \infty}$ and $\displaystyle{\lim_{x \to \infty}(-35x^2) = -35 \lim_{x \to \infty}x^2 = -\infty.}$ What then is the limit as $x \to \infty$ of a polynomial like $f(x) = x^3 -35x^2 + 16x,$ where the limit of the first and third terms go to $+\infty$ but the limit of the second term goes to $-\infty$? Is the limit of the whole function $+\infty,$ or $-\infty$, or 0, or something else?

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

Consider the function $f(x) = x^3 -35x^2 + 16x.$

What do you think its limit is as $x \to \infty$?

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, $x^3,$ 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 $g(x) =x^3$ 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.

Graphs of $f(x) = x^3 -35x^2 + 16x$ and $g(x) = x^3$ versus *x*

to examine behavior as $x \to \infty$

to examine behavior as $x \to \infty$

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:

\begin{array}{lllll}

f(x)&= & x^3 & -\,35x^2 &+ \,16x \\[8px]
f(10\,000) &= & (10\,000)^3 & -\,35(10\,000)^2 &+ \,16(10\,000)\quad \\[8px]
&= &1.0 \times 10^{12} &- \,3.5 \times 10^9 &+ \,1.6 \times 10^5 \\[8px]
\end{array}

Looking term-by-term, we see that at $x=10\,000$ the $x^3$ term is of order $10^{12},$ while the second, next-largest term is only of order $10^{9},$ so an effect approximately $1\,000$-times as small. The third term, $16x,$ is much smaller still: of order $10^{5}$ it is not a small number, but is rather insignificant compared to the first term’s size of $10^{12}.$

Furthermore, remember that $x \to \infty$ means *x* can be as large as we’d like. So if we consider an even larger input value, say $x=10^{20},$ then we have

\begin{array}{lllll}

f(x)&= & x^3 & -\,35x^2 &+ \,16x \\[8px]
f\left(10^{20} \right) &= & \left(10^{20} \right)^3 & -\,35\left(10^{20} \right)^2 &+ \,16\left(10^{20} \right)\quad \\[8px]
&= &1.0 \times 10^{60} &- \,3.5 \times 10^{41} &+ \,1.6 \times 10^{21} \\[8px]
\end{array}

Here the second term is approximately $10^{19}$-times as small as the first term, and the third terms is another $10^{20}$-times as small.

This is all to say that as *x* grows larger, the term with the largest power $\left(x^3 \right)$ dominates over the other terms with smaller powers ($-35x^2$ and $16x$). Hence *when thinking about the limit as* $x \to \infty,$ we can focus *only* on that largest term that dominates, since its limit is the same as the limit as the original function:

\[\bbox[5px,border:2px solid green]{x^3} – 35 x^2 + 16x \, \xrightarrow{\text{as $x$ grows large}} \, x^3 \]
Then simply recall that $\displaystyle{\lim_{x \to \infty}x^3 = \infty}$, which lets you quickly reason that

\begin{align*}

\lim_{x \to \infty} \left(x^3 – 35 x^2 + 16x \right) &= \lim_{x \to \infty}x^3 \\[8px]
&= \infty \quad \cmark

\end{align*}

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 $x^3$:

\[x^3 – 35 x^2 + 16x = x^3 \left( 1 – \frac{35}{x} + \frac{16}{x^2}\right)\]
Then recall that $\displaystyle{\lim_{x \to \infty}\frac{1}{x} = 0}$ and $\displaystyle{\lim_{x \to \infty}\frac{1}{x^2} = 0}$:

\begin{align*}

\lim_{x \to \infty} \left(x^3 – 35 x^2 + 16x \right) &= \lim_{x \to \infty}\left[x^3 \left( 1 – \frac{35}{x} + \frac{16}{x^2}\right) \right] \\[8px]
&= \left[\lim_{x \to \infty}x^3 \right]\left[\lim_{x \to \infty}1 -35\lim_{x \to \infty}\frac{1}{x} + 16\lim_{x \to \infty}\frac{1}{x^2} \right] \\[8px]
&= \left[\lim_{x \to \infty}x^3 \right]\left[\lim_{x \to \infty}1 -35\cancelto{0}{\lim_{x \to \infty}\frac{1}{x}} + 16\cancelto{0}{\lim_{x \to \infty}\frac{1}{x^2}} \right] \\[8px]
&= \left[\lim_{x \to \infty}x^3\right][1] \qquad \left[\text{Recall }\lim_{x \to \infty}x^3 = \infty \right]\\[8px]
&= \infty \quad \cmark

\end{align*}

And that’s our proof that the limit of this polynomial is $\infty,$ confirming our informal reasoning.

For instance, for the function above $f(x) = x^3 – 35 x^2 + 16x,$ we need only consider the $x^3$ term for our reasoning.

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

For quick informal reasoning, use dominance and just look at the largest term:

\[\lim_{x \to \infty} \left( Ax^N + \text{ (smaller terms)}\right) = \lim_{x \to \infty} A x^N \] and then use what you know about that limit from the preceding screens.

\[\lim_{x \to \infty} \left( Ax^N + \text{ (smaller terms)}\right) = \lim_{x \to \infty} A x^N \] 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 $\displaystyle{ \lim_{x \to \infty}x^n }$ and $\displaystyle{ \lim_{x \to \infty}\dfrac{1}{x^n} }.$

You’ll be able to practice these approaches below. Before then, let’s extend our ideas to apply to

$x \to -\infty$, using the same function as above. Since the reasoning regarding dominance is largely the same, we’ll proceed more quickly.

What is $\displaystyle{\lim_{x \to -\infty} \left( x^3 -35x^2 + 16x\right)}$?

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

Graphs of $f(x) = x^3 -35x^2 + 16x$ and $g(x) = x^3$ versus *x*

to examine behavior as $x \to -\infty$

to examine behavior as $x \to -\infty$

*Informal, quick reasoning*

As above, when thinking about the limit we can focus only on the largest term that dominates:

\[\bbox[5px,border:2px solid green]{x^3} – 352 x^2 + 16x \, \xrightarrow{\text{as $x$ grows large}} \, x^3 \]
and then recall that $\displaystyle{\lim_{x \to -\infty}x^3 = -\infty}$, and so

\begin{align*}

\lim_{x \to -\infty} \left(x^3 – 35 x^2 + 16x \right) &= \lim_{x \to -\infty}x^3 \\[8px]
&= -\infty \quad \cmark

\end{align*}

*Formal development of the limit*

Also as above, we first factor out the largest-power term that dominates, and then recall that $\displaystyle{\lim_{x \to -\infty}\frac{1}{x} = 0}$ and $\displaystyle{\lim_{x \to -\infty}\frac{1}{x^2} = 0}$:

\begin{align*}

\lim_{x \to -\infty} \left(x^3 – 35 x^2 + 16x \right) &= \lim_{x \to -\infty}\left[x^3 \left( 1 – \frac{35}{x} +

\frac{16}{x^2}\right) \right] \\[8px]
&= \left[\lim_{x \to -\infty}x^3 \right]\left[\lim_{x \to -\infty}1 -35\lim_{x \to -\infty}\frac{1}{x} + 16\lim_{x \to -\infty}\frac{1}{x^2} \right] \\[8px]
&= \left[\lim_{x \to -\infty}x^3 \right]\left[\lim_{x \to -\infty}1 -35\cancelto{0}{\lim_{x \to -\infty}\frac{1}{x}} + 16\cancelto{0}{\lim_{x \to -\infty}\frac{1}{x^2}} \right] \\[8px]
&= \left[\lim_{x \to -\infty}x^3\right][1] \qquad \left[\text{Recall }\lim_{x \to -\infty}x^3 = -\infty \right]\\[8px]
&= -\infty \quad \cmark

\end{align*}

For quick informal reasoning:

\[\lim_{x \to -\infty} \left( Ax^N + \text{ (smaller terms)}\right) = \lim_{x \to -\infty} A x^N \] For formal development of the limit,**factor out the largest-power term** and then use what you know about the resulting limits for $\displaystyle{ \lim_{x \to -\infty}x^n }$ and $\displaystyle{ \lim_{x \to -\infty}\dfrac{1}{x^n} }.$

\[\lim_{x \to -\infty} \left( Ax^N + \text{ (smaller terms)}\right) = \lim_{x \to -\infty} A x^N \] For formal development of the limit,

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 $x \to \pm \infty$ 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.

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- When considering the limit on a polynomial as $x \to \infty$ or $x \to -\infty,$ 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.* - 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:

\[\lim_{x \to \infty} \left( Ax^N + \text{ (smaller terms)}\right) = \lim_{x \to \infty} A x^N \] \[\lim_{x \to -\infty} \left( Ax^N + \text{ (smaller terms)}\right) = \lim_{x \to -\infty} A x^N \] - 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 $\lim_{x \to \infty}.$ We’re waiting for you there!