Are you working with Limit at Infinity problems in Calculus? Let’s break ’em down and develop your understanding so you can solve them routinely for yourself.

I.   How to think about “going to infinity”

II.  $\displaystyle{\lim_{x \to \infty} \frac{1}{x^n}}$  and $\displaystyle{\lim_{x \to \,-\infty} \frac{1}{x^n}}$

We’re taking $n>0$ in this discussion, so we’re looking at functions like $\dfrac{1}{x^{1/3}}$,  or $\dfrac{1}{x^2}$,  or $\dfrac{1}{x^5}.$
A. Positive direction: $\displaystyle{\lim_{x \to \infty}\frac{1}{x^n}}$

B. Negative direction: $\displaystyle{\lim_{x \to \,-\infty}\frac{1}{x^n}}$

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III.  $\displaystyle{\lim_{x \to \infty} x^n }$  and $\displaystyle{\lim_{x \to \,-\infty} x^n }$

Quick vocabulary note: When we write that a limit “equals” $\infty$ or $-\infty$, remember that really we mean that the function grows and Grows and GROWS forever, without ever reaching a particular value. (Again: infinity is not a particular place or some “magic number;” instead it is a process of never-ending growth.) Hence in these cases the limit does not exist (DNE).
In some situations, like a multiple-choice question (on the AP Calculus exam, among other places), the correct answer may be listed as “does not exist” or “DNE,” or “undefined,” rather than “$\infty$” or “$-\infty$”.

IV. Polynomials: The highest power dominates

What happens as x grows and Grows and GROWS when we have a polynomial, which by definition has different powers of x in it? Let’s look at a specific example to investigate.

Once you’re comfortable with this reasoning, you’ll be able to answer these questions correctly almost immediately upon reading them.

Let’s apply this reasoning (along with more formal proof-type reasoning) to the same function, now looking at what happens as x grows large in the negative direction. The same approach applies:

The upshot: Find the term in the polynomial with the largest power of x. As x grows and Grows and GROWS (in either the positive or negative direction), that term dominates, and so the limit of the polynomial is the same as the limit of that single term.

$$\bbox[yellow,5px]{\lim_{x \to \infty} \left( Ax^N + \text{ (smaller terms)}\right) = \lim_{x \to \infty} A x^N }$$

$$\bbox[yellow,5px]{\lim_{x \to\, -\infty} \left( Ax^N + \text{ (smaller terms)}\right) = \lim_{x \to\, -\infty} A x^N }$$

If you want to prove your result more rigorously, start by factoring out x-to-the-largest-power. Then proceed to find the limit as shown in the examples above.

V. Fraction of polynomials (aka “Rational Functions”)

If you have one polynomial divided by another polynomial (technically called a rational function), then:

Step 1. Identify the term in the numerator with the highest power, and the term in the denominator with the highest power. You can then ignore all of the other terms just as we did for the polynomials in the preceding section.

Step 2. Compare the power in the numerator to the power in the denominator. There are three possibilities: (1) the highest power in the numerator is the same as that in the denominator; (2) the highest power in the denominator is greater than that in the numerator; (3) the highest power in the numerator is greater that in the denominator. Let’s consider each:

• Possibility 1: Equal powers. If the power in the numerator is the same as the power in the denominator, then the limit is the ratio of the coefficients of those two terms. This rule holds for both $x \to \infty$  and $x \to \, -\infty.$
$$\bbox[yellow,5px]{\lim_{x \to \infty}\frac{Ax^N + \text{ (smaller terms)}}{Bx^N + \text{ (smaller terms)}} = \lim_{x \to \infty}\frac{Ax^N}{Bx^N} = \frac{A}{B}}$$
$$\bbox[yellow,5px]{\lim_{x \to\, -\infty}\frac{Ax^N + \text{ (smaller terms)}}{Bx^N + \text{ (smaller terms)}} = \lim_{x \to \, -\infty}\frac{Ax^N}{Bx^N} = \frac{A}{B}}$$

Conceptually, the numerator and denominator are growing at the same rate, as modified only by the coefficients of those largest terms.

Example 3

Find $\displaystyle{\lim_{x \to \infty} \frac{2x^4 + 5x^2 + 18x – 9}{5x^4 + 47x^2 + 138}}.$

Solution.
\[ \begin{align*}\lim_{x \to \infty} \frac{2x^4 + 5x^2 + 18x – 9}{5x^4 + 47x^2 + 138} &= \lim_{x \to \infty} \frac{2x^4}{5x^4} \\[8px] &= \frac{2}{5} \quad \cmark
\end{align*} \] That’s it; you’re done.

Open to develop this result with more formal reasoning

The “trick” to remember for these problems is to (1) identify the largest power in the denominator, and then (2) divide every term in the expression by x-to-that-power.

For example, in this problem the largest power in the denominator is $x^4$, so we first divide each and every term by $x^4.$
\[ \begin{align*}
\lim_{x \to \infty} \frac{2x^4 + 5x^2 + 18x – 9}{5x^4 + 47x^2 + 138} &= \lim_{x \to \infty}\dfrac{\dfrac{2x^4}{x^4} + \dfrac{5x^2}{x^4} + \dfrac{18x}{x^4}\, – \dfrac{9}{x^4}}{\dfrac{5x^4}{x^4}\, + \dfrac{47x^2}{x^4} + \dfrac{138}{x^4}} \\[8px] &= \lim_{x \to \infty}\dfrac{2 + \dfrac{5}{x^2} + \dfrac{18}{x^3} \, – \dfrac{9}{x^4}}{5\, – \dfrac{47}{x^2} + \dfrac{138}{x^4}} \\[8px] &= \frac{2 + 0 + 0 – 0}{5 – 0 + 0} \\[8px] &= \frac{2}{5} \quad \cmark
\end{align*} \] Notice that in going from the second to the third line, we used the fact that $\displaystyle{\lim_{x \to \infty}\frac{1}{x} = 0}$,   $\displaystyle{\lim_{x \to \infty}\frac{1}{x^2} = 0}$,   and so forth.

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We can verify the result by looking at the graph of the function. Notice that it approaches the line $y = \dfrac{2}{5}$ as $x \to \infty.$

We can also have a horizontal asymptote as $x \to \, -\infty$, as Example 4 illustrates.

Example 4

Find $\displaystyle{ \lim_{x \to \, -\infty} \frac{-x^2 – 3x – 6}{2x^2 + 8} }.$

Solution.
\[ \begin{align*}
\lim_{x \to \, -\infty} \frac{-x^2 – 3x – 6}{2x^2 + 8} &= \lim_{x \to \, -\infty} \frac{-x^2}{2x^2} \\[8px] &= -\frac{1}{2} \quad \cmark
\end{align*} \]

Open to see how to prove this result

We use the same “trick” as we did in Example 3: (1) identify the largest power in the denominator, and then (2) divide every term in the expression by x-to-that-power.

In this problem, the largest power present in the denominator is $x^2$, so we divide each and every term in the expression by $x^2$:
\[ \begin{align*}
\lim_{x \to \, -\infty} \frac{-x^2 – 3x – 6}{2x^2 + 8} &= \lim_{x \to \, -\infty} \frac{-\dfrac{x^2}{x^2}\, – \dfrac{3x}{x^2}\, – \dfrac{6}{x^2}}{\dfrac{2x^2}{x^2}\, + \dfrac{8}{x^2}} \\[8px] &= \lim_{x \to \, -\infty} \dfrac{-1 -\dfrac{3}{x}- \dfrac{6}{x^2}}{2 + \dfrac{8}{x^2}} \\[8px] &= -\frac{1}{2} \quad \cmark
\end{align*} \] Note that to go from the second to the third line, we used the fact that $\displaystyle{\lim_{x \to\, -\infty}\frac{1}{x} = 0}$ and   $\displaystyle{\lim_{x \to \, -\infty}\frac{1}{x^2} = 0}.$

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We can verify the result by looking at the graph of the function. Notice that the curve $y = f(x)$ approaches the line $y = -\dfrac{1}{2}$ as $x \to \,-\infty$, and so that line is a horizontal asymptote for this function.

• Possibility 2: Denominator dominates. If the largest power in the denominator is greater than the largest power in the numerator, then the limit is 0. This rule holds for both $x \to \infty$  and $x \to \, -\infty.$
$$\bbox[yellow,5px]{\lim_{x \to \infty} \frac{\text{(some polynomial)}}{\text{(polynomial whose highest power > numerator’s)}} = 0 }$$
$$\bbox[yellow,5px]{\lim_{x \to \, -\infty} \frac{\text{(some polynomial)}}{\text{(polynomial whose highest power > numerator’s)}} = 0 }$$

Conceptually, the growth in the denominator “wins out” over the growth in the numerator, meaning the denominator grows large at a faster rate than the numerator does, and so the fraction tends toward zero as x grows and Grows and GROWS in either the positive or negative direction.

Example 5

(a) Find $\displaystyle{\lim_{x \to \infty}\frac{x^3 – 3x^2 + 6x}{2x^6 + x^2 + 9}}.$

(b) Find $\displaystyle{\lim_{x \to \, -\infty}\frac{x^3 – 3x^2 + 6x}{2x^6 + x^2 + 9}}.$

Solution.
(a)
$$\lim_{x \to \infty}\frac{x^3 – 3x^2 + 6x}{2x^6 + x^2 + 9} = \lim_{x \to \infty}\frac{x^3}{2x^6} = 0 \quad \cmark$$
because the highest power in the denominator is greater than the highest power in the numerator. (That’s it; the reasoning is that simple.)

Open to see how to prove this result

We use the same “trick” as we did in Examples 3 & 4 above: (1) identify the largest power in the denominator, and then (2) divide every term in the expression by x-to-that-power. Here the highest power in the denominator is $x^6$, and so we divide each and every term by that power:
\[ \begin{align*}
\lim_{x \to \infty}\frac{x^3 – 3x^2 + 6x}{2x^6 + x^2 + 9} &= \lim_{x \to \infty}\frac{\dfrac{x^3}{x^6} – \dfrac{3x^2}{x^6} + \dfrac{6x}{x^6}}{\dfrac{2x^6}{x^6} + \dfrac{x^2}{x^6} + \dfrac{9}{x^6}} \\[8px] &= \lim_{x \to \infty}\frac{\dfrac{1}{x^3} – \dfrac{3}{x^4} + \dfrac{6}{x^5}}{2 + \dfrac{1}{x^4} + \dfrac{9}{x^6}} \\[8px] &= \frac{0 – 0 + 0}{2 + 0 + 0} \\[8px] &= 0 \quad \cmark
\end{align*} \] Note that to go from the second to the third line, we used the fact that $\displaystyle{\lim_{x \to \infty}\frac{1}{x^3} = 0}$, and   $\displaystyle{\lim_{x \to \infty}\frac{1}{x^4} = 0}$, and so forth.

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We can verify the result by looking at the graph of the function. Notice that the curve $y = f(x)$ approaches the line $y = 0$ as $x \to \infty$, and so that line (the horizontal axis) is a horizontal asymptote for this function.

(b)
$$\lim_{x \to \, -\infty}\frac{x^3 – 3x^2 + 6x}{2x^6 + x^2 + 9} = \lim_{x \to\, -\infty}\frac{x^3}{2x^6} = 0 \quad \cmark$$
because the highest power in the denominator is greater than the highest power in the numerator.

Open to see how to prove this result

We use the same “trick” as we did in Examples 3 & 4 above: (1) identify the largest power in the denominator, and then (2) divide every term in the expression by x-to-that-power. Here the highest power in the denominator is $x^6$, and so we divide each and every term by that power:
\[ \begin{align*}
\lim_{x \to \,-\infty}\frac{x^3 – 3x^2 + 6x}{2x^6 + x^2 + 9} &= \lim_{x \to\, -\infty}\frac{\dfrac{x^3}{x^6} – \dfrac{3x^2}{x^6} + \dfrac{6x}{x^6}}{\dfrac{2x^6}{x^6} + \dfrac{x^2}{x^6} + \dfrac{9}{x^6}} \\[8px] &= \lim_{x \to \,-\infty}\frac{\dfrac{1}{x^3} – \dfrac{3}{x^4} + \dfrac{6}{x^5}}{2 + \dfrac{1}{x^4} + \dfrac{9}{x^6}} \\[8px] &= \frac{0 – 0 + 0}{2 + 0 + 0} \\[8px] &= 0 \quad \cmark
\end{align*} \] Note that to go from the second to the third line, we used the fact that $\displaystyle{\lim_{x \to\, -\infty}\frac{1}{x^3} = 0}$, and   $\displaystyle{\lim_{x \to \,-\infty}\frac{1}{x^4} = 0}$, and so forth.

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We can verify the result by looking at the graph of the function. Notice that the curve $y = f(x)$ approaches the line $y = 0$ as $x \to\, -\infty$, and so that line (the horizontal axis) is a horizontal asymptote for this function.

• Possibility 3: Numerator dominates. If the largest power in the numerator is greater than the largest power in the denominator, then the limit is either $\infty$ or $-\infty$, and hence does not exist (DNE). This rule holds for both $x \to \infty$  and $x \to \, -\infty.$
$$\bbox[yellow,5px]{\lim_{x \to \infty} \frac{\text{(polynomial whose highest power > denominator’s)}}{\text{(some polynomial)}} = \infty \text{ or } -\infty}$$
$$\bbox[yellow,5px]{\lim_{x \to \, -\infty}\frac{\text{(polynomial whose highest power > denominator’s)}}{\text{(some polynomial)}} = \infty \text{ or } -\infty}$$

Conceptually, the growth in the numerator “wins out” over the growth in the denominator, meaning the numerator grows large at a faster rate than the denominator does, and so the fraction just gets bigger and bigger x grows and Grows and GROWS in either the positive or negative direction.

If you need to determine whether the limit is $\infty$ versus $-\infty$, then you have a little more work to do. The steps build on all of the ideas we developed above, as illustrated in the following example.

Over to you:

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