Calculus 1Limits

Limits at Infinity and Horizontal Asymptotes

Compare the growth of the highest powers.

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So far $x$ approached a number. Now let $x$ grow without bound. This describes the long-run behavior of a function and its horizontal asymptotes, which you first met in Asymptotes and Holes. The Limit Laws still apply.

Key idea

$\lim_{x \to \infty} f(x) = L$ means $f(x)$ gets as close to $L$ as we like for all large enough $x$; then $y = L$ is a horizontal asymptote. For a rational function, divide top and bottom by the highest power of $x$ in the denominator. Equal degrees give the ratio of leading coefficients; a lower top gives 0; a higher top gives $\pm\infty$.

Why it works

For large $x$ the highest power dominates. In $\frac{3x^2 - x}{5x^2 + 2}$, dividing every term by $x^2$ gives $\frac{3 - 1/x}{5 + 2/x^2}$. Each term like $\frac{1}{x}$ or $\frac{2}{x^2}$ tends to 0 because a fixed number divided by a huge one is tiny. What survives is $\frac{3}{5}$. The lower-order terms matter near the origin, not in the long run.
$\frac{2x^2+3x}{x^2+1}$ approaches the line $y = 2$ far to the left and far to the right.

Worked example: Equal degrees

Find $\displaystyle\lim_{x \to \infty} \frac{3x^2 - x}{5x^2 + 2}$.
  1. Divide by $x^2$: $\frac{3 - \frac{1}{x}}{5 + \frac{2}{x^2}}$.
  2. The small terms vanish, leaving $\frac{3}{5}$.

Answer: $\frac{3}{5}$, so $y = \frac{3}{5}$ is a horizontal asymptote.

Worked example: A root and a negative direction

Find $\displaystyle\lim_{x \to -\infty} \frac{\sqrt{4x^2 + 1}}{x - 3}$.
  1. Here $\sqrt{x^2} = \lvert x \rvert$, and for negative $x$, $\lvert x \rvert = -x$.
  2. Divide top and bottom by $\lvert x \rvert = -x$: top $\sqrt{4 + \frac{1}{x^2}} \to 2$; bottom $\frac{x - 3}{-x} = -1 + \frac{3}{x} \to -1$.
  3. The limit is $\frac{2}{-1}$.

Answer: $-2$ (and $+2$ as $x \to \infty$)

Worked example: Infinity minus infinity

Find $\displaystyle\lim_{x \to \infty} \left(\sqrt{x^2 + x} - x\right)$.
  1. Both terms grow without bound, so the form is "$\infty - \infty$", which is not automatically 0.
  2. Multiply by the conjugate: $\frac{(x^2 + x) - x^2}{\sqrt{x^2 + x} + x} = \frac{x}{\sqrt{x^2 + x} + x}$.
  3. Divide by $x$: $\frac{1}{\sqrt{1 + 1/x} + 1} \to \frac{1}{2}$.

Answer: $\frac{1}{2}$

Long-run behavior of a rational function, by degree.
Degrees of top and bottom$\lim_{x \to \pm\infty}$ of the rational functionHorizontal asymptote
top lower than bottom0$y = 0$
equalratio of leading coefficients$y = \frac{a}{b}$
top higher$\infty$ or $-\infty$ (check the signs)none

A function can have different limits at $\infty$ and $-\infty$, as the second example shows. That happens with roots of $x^2$ and with exponentials, so always check the direction.

Common mistake

Treating $\frac{\infty}{\infty}$ as 1 or $\infty - \infty$ as 0. Counterexamples: $\frac{2x}{x} \to 2$, and the last example gives $\frac{1}{2}$, not 0. These forms only say "more work needed." Check by plugging in a large number such as $x = 10^6$.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Sources and further reading

  • OpenStax Calculus Volume 1, 4.6 Limits at Infinity and Asymptotes
  • OpenStax Calculus Volume 1, 2.3 The Limit Laws
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