Limits at infinity and asymptotes

So far the input has approached a finite point. Let it run off instead: $x \to \infty$ asks what the function settles down to far out, and the answer is a horizontal asymptote when there is one.

This is the same limit idea with a different approach, and the technique is different too — nothing to factor and cancel, because nothing is $\tfrac{0}{0}$. The competition is between things growing.

The one move: divide by the dominant term

For a ratio of polynomials, divide top and bottom by the highest power of $x$ appearing anywhere in the fraction. Every term then either survives as a constant or dies as $\tfrac{c}{x^k} \to 0$.

$$\lim_{x \to \infty}\frac{3x^2 + 5x - 1}{2x^2 - 7} = \lim_{x \to \infty}\frac{3 + \tfrac{5}{x} - \tfrac{1}{x^2}}{2 - \tfrac{7}{x^2}} = \frac{3}{2}$$

Those three outcomes are the only ones, and a degree comparison predicts which without doing any work:

Roots need care about sign

$\sqrt{x^2} = |x|$, not $x$. Going to $+\infty$ that distinction is invisible; going to $-\infty$ it flips the answer’s sign.

$$\lim_{x \to \infty}\frac{\sqrt{4x^2+1}}{x} = 2 \qquad\text{but}\qquad \lim_{x \to -\infty}\frac{\sqrt{4x^2+1}}{x} = -2$$

Both are verified above, and they differ. A student who writes $\sqrt{x^2} = x$ gets the second one wrong and has no way to notice.

Which grows faster

Beyond polynomials, limits at infinity are decided by a ranking. Each of these beats everything to its left, eventually and permanently:

$$\ln x \;\ll\; x^{p} \;\ll\; e^{x} \;\ll\; x!$$

“Eventually” is doing real work in that sentence. $x^{100}$ is vastly larger than $e^x$ for a long time — they cross somewhere past $x \approx 1000$ — and after the crossing the exponential wins by margins that grow without bound. A table of small values would tell you the opposite of the truth.

The middle claim is the useful one: any power divided by an exponential tends to zero. It settles most questions of this kind on sight.

Vertical asymptotes

A horizontal asymptote comes from $x \to \infty$. A vertical one comes from the other kind of limit — a finite point where the function grows without bound.

For a rational function, that is a zero of the denominator which is not also a zero of the numerator. When it is a zero of both, the factor cancels and you get a hole instead.

$$\frac{x-1}{(x-1)(x+2)}$$

has a hole at $x = 1$ and a vertical asymptote at $x = -2$ — the same expression producing both kinds of behaviour, distinguished only by whether the factor cancels.

At $x = 1$ the limit is a perfectly ordinary $\tfrac13$ — the hole is invisible to the limit, which is chapter 1.1’s point arriving again. At $x = -2$ nothing cancels, and there is no limit at all.