AP Calculus AB and BC

Horizontal Asymptote vs Limit at Infinity

Q: Does every limit at infinity give a horizontal asymptote?

Only a finite one does. If the limit is \pm\infty or does not exist, the computation still has an answer but no line is certified.

Q: Do I have to check both directions?

Yes. The limits as x \to \infty and x \to -\infty are separate computations, and they can disagree, as they do for f(x) = \frac{x}{\sqrt{x^2 + 1}} where they give 1 and -1.

These are the same fact in two forms. The limit at infinity is the computation you carry out, and the horizontal asymptote is the line that limit certifies: the graph has the asymptote y equals L exactly when the limit is the finite number L. The computation gives a number, the asymptote is a line.

Horizontal asymptote

Use when: The question asks for the equation of a line, for end behaviour to sketch, or for how many asymptotes a graph has.

Limit at infinity

Use when: The question says evaluate, or you are doing the algebra: comparing degrees, dividing by the highest power, or applying L'Hopital's rule.

Side by side

	Horizontal asymptote	Limit at infinity
What it is	A line, $y = L$	A number, $L$
How you write it	The horizontal asymptote is $y = 2$	$\lim_{x \to \infty} f(x) = 2$
Exists when	The limit at infinity is a finite number	The function approaches one finite value; otherwise the limit is $\pm\infty$ or does not exist
How many	At most two, one per direction	One per direction, computed separately
Common trap	Believing the graph can never touch it	Reporting the number when the question wanted a line

One is the work and the other is the conclusion. You evaluate the limit as $x$ grows without bound; if the answer is a finite number $L$ , the graph has a horizontal asymptote in that direction. If the limit is infinite or fails to exist, there is no horizontal asymptote that way, no matter what the picture suggests.

\text{For finite } L: \quad \lim_{x \to \infty} f(x) = L \iff y = L \text{ is a horizontal asymptote as } x \to \infty

Because the limit is a statement about the tail only, crossing is allowed. The function $f(x) = \frac{\sin x}{x}$ meets $y = 0$ at every nonzero multiple of $\pi$ and still has $y = 0$ as its horizontal asymptote, since the oscillation is squeezed to zero as $x$ grows.

When the limit is infinite

For $f(x) = \frac{x^2 + 1}{x}$ the limit at infinity is $\infty$ , so there is no horizontal asymptote. The end behaviour still has a shape, following the slant line $y = x$ , but that is a different feature and a different question.

Frequently asked questions

Can a graph cross its horizontal asymptote?

Yes, any number of times. The asymptote reports long-run behaviour only, so crossings near the origin, or infinitely many of them as with $\frac{\sin x}{x}$ , are ordinary.

Does every limit at infinity give a horizontal asymptote?

Only a finite one does. If the limit is $\pm\infty$ or does not exist, the computation still has an answer but no line is certified.

Do I have to check both directions?

Yes. The limits as $x \to \infty$ and $x \to -\infty$ are separate computations, and they can disagree, as they do for $f(x) = \frac{x}{\sqrt{x^2 + 1}}$ where they give $1$ and $-1$ .

In the CED: Unit 1: Limits and Continuity