## Limits at Infinity: What You Need to Know

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”

• When you see $\displaystyle{\lim_{x \to \, \infty}}$, think:

“The limit as *x* grows and grows, and Grows, and GROWS, …, in the positive direction, forever.”

• When you see $\displaystyle{\lim_{x \to \, -\infty}}$, think:

“The limit as *x* grows and grows, and Grows, and GROWS, …, in the negative direction, forever.”

**never-ending process of growth**.

### 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: $\displaystyle{\lim_{x \to \infty}\frac{1}{x^n}}$

• $\bbox[yellow,5px]{\displaystyle{\lim_{x \to \infty}\frac{1}{x^n} = 0} \text{ for } n > 0}$

For example, $\displaystyle{\lim_{x \to \infty}\frac{1}{x} = 0}$: As *x* grows larger and Larger as we move along the *x*-axis to the right, forever, $y=\dfrac{1}{x}$ gets closer and closer to 0. Hence the “limit at infinity” is 0.

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

• $\bbox[yellow,5px]{\displaystyle{\lim_{x \to \,-\infty}\frac{1}{x^n} = 0} \text{ for } n > 0}$

For example, $\displaystyle{\lim_{x \to\, -\infty}\frac{1}{x} = 0}$: As *x* grows larger and Larger in the negative direction as we move along the *x*-axis to the left, forever, $y=\dfrac{1}{x}$ gets closer and closer to 0. Hence the “limit at negative infinity” is 0.

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

A. Positive direction: $\displaystyle{\lim_{x \to \infty}x^n}$

• $\bbox[yellow,5px]{\displaystyle{\lim_{x \to \infty}x^n = \infty} \text{ for all } n > 0}$

For example, $\displaystyle{\lim_{x \to \infty}x^2 = \infty}$: As we move along the *x*-axis to the right, forever, $y=x^2$ grows and Grows upward, in the positive *y*-direction, forever. Hence we write the limit as $\infty.$

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

• $\bbox[yellow,5px]{\displaystyle{\lim_{x \to \,-\infty}x^n = \infty} \text{ for all even } n > 0 }$

since a negative-number to an even power (2, 4, 6, …) is positive.

For example, $\displaystyle{\lim_{x \to \, -\infty}x^2 = \infty}$: As we move along the *x*-axis to the left, forever, $y=x^2$ grows and Grows upward, in the positive *y*-direction, forever. Hence we write the limit as $\infty.$

• $\bbox[yellow,5px]{\displaystyle{\lim_{x \to \,-\infty}x^n = -\infty} \text{ for all odd }n > 0}$

since a negative-number to an odd power (1, 3, 5, …) is negative.

For example, $\displaystyle{\lim_{x \to \, -\infty}x^3 = -\infty}$: As we move along the *x*-axis to the left, forever, $y=x^3$ grows and Grows downward, in the negative *y*-direction, forever. Hence we write the limit as $-\infty.$

**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.

*Solution.*

Your intuition might be that, as *x* grows and Grows and GROWS, forever, the $x^3$ term dominates — meaning that since it is the largest power, it grows at the fastest rate and “overwhelms” all of the other terms. That intuition is correct, as we can see from the following graphs.

Let’s first compare $f(x) = x^3 – 352 x^2 + 16x$ to $x^3$ for $0 \lt x \lt 300.$ They don’t look very similar.

But if we zoom out to $0 \lt x \lt 3,000$, we see that the graph of *f(x)* is starting to approach that of $x^3$ for large values of *x*.

And if we zoom out further to $0 \lt x \lt 30,000$, we see that for large values of *x*, the two graphs practically coincide.

The key point is that as *x* grows large, the term with the highest power $\left(x^3 \right)$ dominates. We can ignore all of the other terms since they have little effect:

$$\bbox[5px,border:2px solid green]{x^3} – 352 x^2 + 16x \, \xrightarrow{\text{as $x$ grows large}} \, x^3 $$

And since we know $\displaystyle{\lim_{x \to \infty}x^3 = \infty}$, we can conclude

\[ \begin{align*}

\lim_{x \to \infty} \left(x^3 – 352 x^2 + 16x \right) &= \lim_{x \to \infty}x^3 \\[8px]
&= \infty \quad \cmark

\end{align*} \]

$$x^3 – 352 x^2 + 16x = x^3 \left( 1 – \frac{352}{x} + \frac{16}{x^2}\right) $$

Now recall that $\displaystyle{\lim_{x \to \infty}\frac{1}{x} = 0}$ and $\displaystyle{\lim_{x \to \infty}\frac{1}{x^2} =0}.$ Hence

$$\lim_{x \to \infty}\left( 1 – \frac{352}{x} + \frac{16}{x^2}\right) = 1 $$

Thus

\[ \begin{align*}

\lim_{x \to \infty} \left(x^3 – 352 x^2 + 16x \right) &= \lim_{x \to \infty}\left[ \left( x^3\right) \left( 1 – \frac{352}{x} + \frac{16}{x^2}\right)\right] \\[8px] &= \left(\lim_{x \to \infty} x^3\right) \lim_{x \to \infty}\left( 1 – \frac{352}{x} + \frac{16}{x^2}\right) \\[8px] &= \left(\lim_{x \to \infty}x^3 \right)(1) \\[8px] &= \infty \quad \cmark

\end{align*} \] since we know that $\displaystyle{\lim_{x \to \infty}x^3 = \infty}.$

The approach of first factoring out

*x*-to-the-largest-power, and then examining the remaining terms, will always work to show that the highest power determines what the polynomial does as $x \to \pm \infty.$

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:

*Solution.*

Conceptual-reasoning quickly leads to the correct answer: The largest term in $f(x) = x^3 – 352 x^2 + 16x$ is $x^3$, and we know that $\displaystyle{\lim_{x \to \, -\infty}x^3 = -\infty}.$ Hence $\displaystyle{\lim_{x \to \, -\infty} \left(x^3 – 352 x^2 + 16x \right) = -\infty} \quad \cmark.$

A quick look at the graph for the function confirms this reasoning.

$$x^3 – 352 x^2 + 16x = x^3 \left( 1 – \frac{352}{x} + \frac{16}{x^2}\right) $$

Now recall that $\displaystyle{\lim_{x \to \,-\infty}\frac{1}{x} = 0}$ and $\displaystyle{\lim_{x \to \,-\infty}\frac{1}{x^2} =0}.$ Hence

$$\lim_{x \to \,-\infty}\left( 1 – \frac{352}{x} + \frac{16}{x^2}\right) = 1 $$

Thus

\[ \begin{align*}

\lim_{x \to \, -\infty} \left(x^3 – 352 x^2 + 16x \right) &= \lim_{x \to \, -\infty}\left[ \left( x^3\right) \left( 1 – \frac{352}{x} + \frac{16}{x^2}\right)\right] \\[8px] &= \left(\lim_{x \to \, -\infty} x^3\right) \lim_{x \to \,-\infty}\left( 1 – \frac{352}{x} + \frac{16}{x^2}\right) \\[8px] &= \left(\lim_{x \to \,- \infty}x^3 \right)(1) \\[8px] &= -\infty \quad \cmark

\end{align*} \] since we know that $\displaystyle{\lim_{x \to \, -\infty}x^3 = -\infty}.$

**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.

*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.

**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.

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

**horizontal asymptote**: If $\displaystyle{\lim_{x \to \infty}f(x) = L}$, where

*L*is a real number, then the graph of

*f(x)*gets arbitrarily close to the line $y = L$ as $x \to \infty.$ That line $y = L$ is called a horizontal asymptote of the graph of

*f*.

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

*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*} \]

**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 $\displaystyle{\lim_{x \to \, -\infty}\frac{1}{x^2} = 0}.$

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.

**(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.)

**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.

**(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.

**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.

• __ 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.

**(a)**Find $\displaystyle{\lim_{x \to \infty} \frac{x^5+3x^3+2x}{3x^4+8}}.$

**(b)**Find $\displaystyle{\lim_{x \to \, -\infty} \frac{x^5+3x^3+2x}{3x^4+8}}.$

*Solution.*

**(a)**

\[ \begin{align*}

\lim_{x \to \infty} \frac{x^5+3x^3+2x}{3x^4+8} &= \lim_{x \to \infty}\frac{x^5}{3x^4} \\[8px]
&= \lim_{x \to \infty}\frac{x}{3} \\[8px]
&= \infty \quad \cmark

\end{align*} \] The second line in the solution shows that the function approaches $\dfrac{x}{3}$ as *x* grows large, matching what the graph shows. The “limit at infinity” is (positive) $\infty$ because the function grows in the positive *y*-direction forever as *x* grows larger and Larger in the positive direction.

**(b)**

\[ \begin{align*}

\lim_{x \to \, -\infty} \frac{x^5+3x^3+2x}{3x^4+8} &= \lim_{x \to\, -\infty}\frac{x^5}{3x^4} \\[8px]
&= \lim_{x \to\, -\infty}\frac{x}{3} \\[8px]
&= -\infty \quad \cmark

\end{align*} \] The second line in the solution shows that the function approaches $\dfrac{x}{3}$ as *x* grows large, matching what the graph shows. The “limit at negative infinity” is negative $\infty$ because the function grows in the negative *y*-direction forever as *x* grows larger and Larger in the negative direction.

By the way, the line $y = \dfrac{x}{3}$ is a **tilted asymptote** (in both the positive and negative directions) for this function, since the graph of the function gets arbitrarily close to that line as *x* increases (in both directions).

Over to you:

- Or what questions do you have?
- What tips do you have to share about solving Limit at Infinity problems?
- How is your study of Calculus going so far?

Please comment below!

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