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.
Idea clave
$\lim_{x \to \infty} f(x) = L$ 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 $\pm\infty$.
Por qué funciona
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.
Ejemplo resuelto: Equal degrees
Find $\displaystyle\lim_{x \to \infty} \frac{3x^2 - x}{5x^2 + 2}$.
- Divide by $x^2$: $\frac{3 - \frac{1}{x}}{5 + \frac{2}{x^2}}$.
- The small terms vanish, leaving $\frac{3}{5}$.
Respuesta: $\frac{3}{5}$, so $y = \frac{3}{5}$ is a horizontal asymptote.
Ejemplo resuelto: A root and a negative direction
Find $\displaystyle\lim_{x \to -\infty} \frac{\sqrt{4x^2 + 1}}{x - 3}$.
- Here $\sqrt{x^2} = \lvert x \rvert$, and for negative $x$, $\lvert x \rvert = -x$.
- Divide top and bottom by $\lvert x \rvert = -x$: top $\sqrt{4 + \frac{1}{x^2}} \to 2$; bottom $\frac{x - 3}{-x} = -1 + \frac{3}{x} \to -1$.
- The limit is $\frac{2}{-1}$.
Respuesta: $-2$ (and $+2$ as $x \to \infty$)
Ejemplo resuelto: Infinity minus infinity
Find $\displaystyle\lim_{x \to \infty} \left(\sqrt{x^2 + x} - x\right)$.
- Both terms grow without bound, so the form is "$\infty - \infty$", which is not automatically 0.
- Multiply by the conjugate: $\frac{(x^2 + x) - x^2}{\sqrt{x^2 + x} + x} = \frac{x}{\sqrt{x^2 + x} + x}$.
- Divide by $x$: $\frac{1}{\sqrt{1 + 1/x} + 1} \to \frac{1}{2}$.
Respuesta: $\frac{1}{2}$
| Degrees of top and bottom | $\lim_{x \to \pm\infty}$ of the rational function | Horizontal asymptote |
|---|---|---|
| top lower than bottom | 0 | $y = 0$ |
| equal | ratio 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.
Error común
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 $x = 10^6$.