Limits at Infinity

A limit at infinity is simply where a function ends up if you keep going right (or left) along the graph forever. It's how you describe long-run behavior: where a temperature settles, the top speed of a falling object, the most an enzyme can do. The key fact $\lim_{x\to\infty}\frac{1}{x^r} = 0$ looks technical, but it's just $\frac{1}{x^r}$ (one divided by a big power of $x$) and $\to 0$ (gets tiny): one divided by a million, then a billion, is close to nothing. If $f(x)$ settles toward a number $L$, the graph flattens out along the line $y = L$ (a horizontal asymptote). Everything else in this lesson is rewriting a function so that pieces like $\frac{1}{x^r}$ show up.

Polynomials follow their leading term. Take $2x^3 - 100x^2$. At $x = 10$ the $-100x^2$ wins easily, but factor out $x^3$ and you get $x^3\left(2 - \frac{100}{x}\right)$. As $x$ grows, $\frac{100}{x} \to 0$, so the thing in parentheses heads to $2$ and the whole expression goes to $\infty$. The highest power always wins in the end.

Rational functions work the same way. Divide the top and bottom by the highest power of $x$ in the denominator. Every smaller term turns into something like $\frac{3}{x}$ and fades to $0$, and you're left comparing leading terms:

It makes sense once you see it: if the bottom grows faster, the fraction shrinks; if they grow at the same rate, only their leading coefficients matter; and if the top grows faster, the fraction blows up.

A common mistake. With square roots, it's tempting to write $\sqrt{x^2} = x$ and move on. That's only true for $x \ge 0$. In general $\sqrt{x^2} = |x|$, which is $-x$ when $x$ is negative. So as $x \to -\infty$, factoring $x^2$ out of a square root brings out $-x$, not $x$, and forgetting this flips the sign of your answer.

Long-run behavior is often what matters: the steady-state temperature of an engine, the eventual speed of a falling object with drag, the average cost per unit at high volume.