Limits at Infinity

What happens to a function in the long run, as $x$ gets huge? A population model might level off, a cooling cup of coffee approaches room temperature, and a cost per item might settle at some value. Limits at infinity describe this end behaviour. When a function levels off at a height $L$, the line $y = L$ is a horizontal asymptote.

Key ideas

What a limit at infinity means

$$\lim_{x \to \infty} f(x) = L$$

means $f(x)$ gets as close to $L$ as you like when $x$ is large enough. Then $y = L$ is a horizontal asymptote. The same idea works as $x \to -\infty$, and a function can have a different horizontal asymptote at each end.

A graph can cross a horizontal asymptote. The asymptote only describes what happens far to the left or right.

$y = \dfrac{2x^2 + 3x}{x^2 + 1}$ approaches $y = 2$ at both ends, and crosses it once along the way.

The basic fact

For any positive power $n$,

$$\lim_{x \to \pm\infty} \frac{1}{x^n} = 0$$

(Dividing a fixed number by a huge number gives something tiny.) For $x \to -\infty$, $n$ should be a power for which $x^n$ is defined for negative $x$, like a whole number.

Rational functions: compare the degrees

To find $\displaystyle\lim_{x \to \pm\infty} \frac{p(x)}{q(x)}$, divide every term by the highest power of $x$ in the denominator. Then each leftover term with $x$ in its denominator goes to $0$. The result depends only on the leading terms:

Degrees	Limit as $x \to \pm\infty$	Horizontal asymptote
top degree less than bottom	$0$	$y = 0$
degrees equal	ratio of the leading coefficients	$y = \dfrac{a}{b}$
top degree greater than bottom	$\infty$ or $-\infty$	none

When the top degree is bigger, the sign of the infinite limit comes from the leading terms (see Example 2).

Exponentials and logarithms

$$\lim_{x \to \infty} e^x = \infty, \qquad \lim_{x \to -\infty} e^x = 0, \qquad \lim_{x \to \infty} e^{-x} = 0, \qquad \lim_{x \to \infty} \ln x = \infty$$

So $y = e^x$ has the horizontal asymptote $y = 0$ on the left only. $\ln x$ grows forever, just very slowly.

Which grows fastest?

As $x \to \infty$, exponentials beat powers, and powers beat logarithms:

$$\ln x \ \ll\ x^n\ \ (n \gt 0) \ \ll\ e^x$$

This means, for example, $\dfrac{x^{10}}{e^x} \to 0$ and $\dfrac{\ln x}{x} \to 0$. You’ll be able to prove these with L’Hôpital’s rule later.

Square roots: careful with negative x

$\sqrt{x^2} = |x|$, not $x$. When $x \to -\infty$, $x$ is negative, so $\sqrt{x^2} = -x$. This is why functions with square roots can have two different horizontal asymptotes (Example 4).

Worked examples

Example 1: Equal degrees

Find $\displaystyle\lim_{x \to \infty} \frac{3x^2 - 5x}{6x^2 + 1}$.

Solution. Divide every term by $x^2$, the highest power in the denominator:

$$\lim_{x \to \infty} \frac{3 - \dfrac{5}{x}}{6 + \dfrac{1}{x^2}} = \frac{3 - 0}{6 + 0} = \frac{1}{2}$$

The degrees are equal, so the answer is the ratio of the leading coefficients, $\tfrac{3}{6}$. The horizontal asymptote is $y = \tfrac{1}{2}$.

Example 2: Unequal degrees

Find $\displaystyle\lim_{x \to -\infty} \frac{4x + 1}{x^2 - 3}$ and $\displaystyle\lim_{x \to \infty} \frac{x^3}{2x^2 + 1}$.

Solution. In the first, the top has lower degree. Divide by $x^2$:

$$\lim_{x \to -\infty} \frac{\dfrac{4}{x} + \dfrac{1}{x^2}}{1 - \dfrac{3}{x^2}} = \frac{0}{1} = 0$$

In the second, the top has higher degree. Divide by $x^2$:

$$\lim_{x \to \infty} \frac{x}{2 + \dfrac{1}{x^2}}$$

The top grows without bound while the bottom approaches $2$, so the limit is $\infty$. There’s no horizontal asymptote.

Example 3: Exponentials

Find the horizontal asymptotes of $f(x) = \dfrac{3e^x + 1}{e^x + 2}$.

Solution. As $x \to \infty$, $e^x$ is huge. Divide top and bottom by $e^x$:

$$\lim_{x \to \infty} \frac{3 + e^{-x}}{1 + 2e^{-x}} = \frac{3 + 0}{1 + 0} = 3$$

As $x \to -\infty$, $e^x \to 0$, so substitute $0$ for $e^x$:

$$\lim_{x \to -\infty} \frac{3e^x + 1}{e^x + 2} = \frac{0 + 1}{0 + 2} = \frac{1}{2}$$

The horizontal asymptotes are $y = 3$ (on the right) and $y = \tfrac{1}{2}$ (on the left).

Example 4: A square root

Find $\displaystyle\lim_{x \to \infty} \frac{\sqrt{4x^2 + 1}}{x + 3}$ and $\displaystyle\lim_{x \to -\infty} \frac{\sqrt{4x^2 + 1}}{x + 3}$.

Solution. Factor $x^2$ out of the root: $\sqrt{4x^2 + 1} = \sqrt{x^2}\sqrt{4 + \dfrac{1}{x^2}} = |x|\sqrt{4 + \dfrac{1}{x^2}}$.

As $x \to \infty$, $|x| = x$:

$$\lim_{x \to \infty} \frac{x\sqrt{4 + \frac{1}{x^2}}}{x\left(1 + \frac{3}{x}\right)} = \frac{\sqrt{4}}{1} = 2$$

As $x \to -\infty$, $|x| = -x$:

$$\lim_{x \to -\infty} \frac{-x\sqrt{4 + \frac{1}{x^2}}}{x\left(1 + \frac{3}{x}\right)} = -\frac{\sqrt{4}}{1} = -2$$

Two different horizontal asymptotes: $y = 2$ and $y = -2$.

Common mistakes

Dividing by the wrong power. Divide by the highest power in the denominator. Dividing by a random power can leave you with $\tfrac{\infty}{\infty}$ again.

Thinking a graph can’t cross its horizontal asymptote. It can (see the figure). Vertical asymptotes can’t be crossed, but horizontal ones describe only the far ends.

Forgetting that √(x²) = |x|. As $x \to -\infty$, $\sqrt{x^2} = -x$. Missing the minus sign gives the wrong asymptote on the left.

Assuming both ends behave the same. Exponential functions and square-root expressions often have different limits at $\infty$ and $-\infty$. Check each end separately.

Mixing up infinite limits and limits at infinity. $\displaystyle\lim_{x \to 2} f(x) = \infty$ (an infinite limit) means a vertical asymptote. $\displaystyle\lim_{x \to \infty} f(x) = 2$ (a limit at infinity) means a horizontal asymptote.

Practice

1. (Warm-up) Find $\displaystyle\lim_{x \to \infty} \frac{7}{x^3}$.

Solution

The denominator grows without bound while the top stays at $7$:

$$\lim_{x \to \infty} \frac{7}{x^3} = 0$$

2. (Warm-up) Find $\displaystyle\lim_{x \to \infty} \frac{5x - 2}{2x + 9}$.

Solution

Equal degrees, so take the ratio of the leading coefficients: $\dfrac{5}{2}$.

$$\lim_{x \to \infty} \frac{5 - \frac{2}{x}}{2 + \frac{9}{x}} = \frac{5}{2}$$

3. (Core) Find the horizontal asymptote of $f(x) = \dfrac{2x^2 + 3x}{x^2 + 1}$, and find where the graph crosses it.

Solution

Equal degrees, leading coefficients $2$ and $1$, so the horizontal asymptote is $y = 2$ (at both ends).

To find where $f(x) = 2$:

$$2x^2 + 3x = 2(x^2 + 1) \quad\Rightarrow\quad 3x = 2 \quad\Rightarrow\quad x = \frac{2}{3}$$

The graph crosses the asymptote at $\left(\tfrac{2}{3}, 2\right)$, as in the figure.

4. (Core) Find $\displaystyle\lim_{x \to -\infty} \frac{3x^3 - x}{1 - 2x^3}$.

Solution

Divide by $x^3$:

$$\lim_{x \to -\infty} \frac{3 - \frac{1}{x^2}}{\frac{1}{x^3} - 2} = \frac{3}{-2} = -\frac{3}{2}$$

5. (Core) Find $\displaystyle\lim_{x \to \infty} (4 - 3e^{-x})$ and $\displaystyle\lim_{x \to -\infty} (4 - 3e^{-x})$.

Solution

As $x \to \infty$, $e^{-x} \to 0$, so the limit is $4 - 0 = 4$.

As $x \to -\infty$, $-x \to \infty$, so $e^{-x} \to \infty$ and $4 - 3e^{-x} \to -\infty$.

So $y = 4$ is a horizontal asymptote on the right only.

6. (Core) Find $\displaystyle\lim_{x \to \infty} \frac{\sqrt{9x^2 + x}}{2x - 1}$ and $\displaystyle\lim_{x \to -\infty} \frac{\sqrt{9x^2 + x}}{2x - 1}$.

Solution

$\sqrt{9x^2 + x} = |x|\sqrt{9 + \dfrac{1}{x}}$.

As $x \to \infty$, $|x| = x$:

$$\lim_{x \to \infty} \frac{x\sqrt{9 + \frac{1}{x}}}{x\left(2 - \frac{1}{x}\right)} = \frac{3}{2}$$

As $x \to -\infty$, $|x| = -x$, so the limit is $-\dfrac{3}{2}$.

7. (Core) A tank holds $200$ L of fresh water. Salt water containing $30$ g of salt per litre flows in at $4$ L/min (and nothing flows out). After $t$ minutes the concentration of salt in the tank is

$$C(t) = \frac{120t}{200 + 4t} \text{ g/L}$$

Find $\displaystyle\lim_{t \to \infty} C(t)$ and explain what it means.

Solution

Equal degrees, so divide by $t$:

$$\lim_{t \to \infty} \frac{120}{\frac{200}{t} + 4} = \frac{120}{4} = 30$$

In the long run, the concentration approaches $30$ g/L, the same as the incoming salt water. (The original fresh water becomes a smaller and smaller fraction of the mix.)

8. (Challenge) Find $\displaystyle\lim_{x \to \infty} \left(\sqrt{x^2 + 6x} - x\right)$.

Solution

This is an $\infty - \infty$ form. Multiply by the conjugate over itself:

$$\begin{aligned} \lim_{x \to \infty} \frac{\left(\sqrt{x^2 + 6x} - x\right)\left(\sqrt{x^2 + 6x} + x\right)}{\sqrt{x^2 + 6x} + x} &= \lim_{x \to \infty} \frac{(x^2 + 6x) - x^2}{\sqrt{x^2 + 6x} + x} \\ &= \lim_{x \to \infty} \frac{6x}{x\sqrt{1 + \frac{6}{x}} + x} \\ &= \lim_{x \to \infty} \frac{6}{\sqrt{1 + \frac{6}{x}} + 1} = \frac{6}{2} = 3 \end{aligned}$$

(For $x \gt 0$, $\sqrt{x^2 + 6x} = x\sqrt{1 + \tfrac{6}{x}}$.)

9. (Challenge) Find $\displaystyle\lim_{x \to \infty} \frac{2^x + 3^x}{3^x + 1}$ and $\displaystyle\lim_{x \to -\infty} \frac{2^x + 3^x}{3^x + 1}$.

Solution

As $x \to \infty$, the fastest-growing term is $3^x$. Divide top and bottom by $3^x$:

$$\lim_{x \to \infty} \frac{\left(\frac{2}{3}\right)^x + 1}{1 + 3^{-x}} = \frac{0 + 1}{1 + 0} = 1$$

(since $\left(\tfrac{2}{3}\right)^x \to 0$ when the base is between $0$ and $1$).

As $x \to -\infty$, both $2^x \to 0$ and $3^x \to 0$:

$$\lim_{x \to -\infty} \frac{2^x + 3^x}{3^x + 1} = \frac{0 + 0}{0 + 1} = 0$$