AP Calculus AB and BC

Limit of sin(x)/x as x Approaches Infinity

Q: Can a graph cross its horizontal asymptote?

Yes, and this function is the standard example. It crosses y = 0 at every nonzero multiple of \pi, infinitely often. The definition of a horizontal asymptote constrains only long-run behavior, so it says the values must close in on 0, not that they must stay on one side of it.

Q: Why is L'Hopital's rule not allowed here?

Its hypotheses require the quotient to be \frac{0}{0} or \frac{\infty}{\infty}, and \sin x approaches neither, since it has no limit as x \to \infty. Applying it anyway gives \frac{\cos x}{1}, which also has no limit, so the rule would suggest the original limit fails to exist. The squeeze theorem shows it does exist and equals 0.

Q: Does the same reasoning work as x approaches negative infinity?

Yes, with one adjustment. For x < 0 dividing by x reverses the inequalities, so the bounds swap places and the trap becomes \frac{1}{x} \le \frac{\sin x}{x} \le -\frac{1}{x}. Both bounds still go to 0, so the limit is 0 and y = 0 is a horizontal asymptote on both ends.

The limit of sin x over x as x approaches infinity is 0. The squeeze theorem settles it: sine always stays between -1 and 1, so the quotient is trapped between -1 over x and 1 over x, and both bounds go to 0. This is a different question from the same expression as x approaches 0, where the limit is 1.

\lim_{x \to \infty} \frac{\sin x}{x} = 0

Settled by the squeeze theorem.

Setting up the squeeze

Only one fact about sine is needed, and it holds for every real input regardless of how the oscillation is behaving.

-1 \le \sin x \le 1

Heading to $\infty$ means we may assume $x > 0$ , and dividing an inequality by a positive number leaves the directions alone.

-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x} \quad (x > 0)

The two outer functions are simple, and both close on the same value.

\lim_{x \to \infty} \left(-\frac{1}{x}\right) = 0 = \lim_{x \to \infty} \frac{1}{x}

With the upper and lower bounds pinned to the same limit, the function between them has nowhere else to go.

The pattern to remember

A bounded numerator over an unbounded denominator goes to 0. Sine, cosine, and any expression trapped between two constants all qualify, which is why $\frac{\cos x}{x}$ and $\frac{\sin(x^{2})}{x}$ have the same limit of 0 with the same one-line argument.

Why substitution and the usual reflexes fail

Substituting produces something worse than an indeterminate form. The denominator is unbounded, but the numerator has no limit at all: $\sin x$ keeps oscillating between $-1$ and 1 forever and never settles, so there is nothing to write in the numerator slot.

That also blocks the quotient law. Splitting a limit of a quotient into a quotient of limits requires both limits to exist, and $\lim_{x \to \infty} \sin x$ does not.

L'Hopital's rule is unavailable for the same reason. The form is not $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , since the numerator does neither. Differentiating anyway is actively misleading.

\lim_{x \to \infty} \frac{\cos x}{1} \quad \text{does not exist}

A student who applied the rule without checking would conclude the original limit fails to exist, which is wrong. When L'Hopital's hypotheses are not met, its output carries no information about the original limit.

The mistake students make

By far the most common error is answering 1, imported from the special trig limit that gets memorized in Unit 1. Same expression, different approach point, different technique, different answer.

Question	What the form is	Value
$\lim_{x \to 0} \frac{\sin x}{x}$	$\frac{0}{0}$	$1$
$\lim_{x \to \infty} \frac{\sin x}{x}$	Bounded over unbounded	$0$

The near-0 case is a genuine indeterminate form where numerator and denominator shrink at matching rates. The at-infinity case is not indeterminate at all, because the numerator never grows. Reading the approach point before selecting a method is what separates the two.

The second surprise is the graph. This function equals 0 at $x = n\pi$ for every nonzero integer $n$ , so it crosses its own horizontal asymptote infinitely many times while the oscillations are damped by the shrinking $\frac{1}{x}$ envelope.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Can a graph cross its horizontal asymptote?

Yes, and this function is the standard example. It crosses $y = 0$ at every nonzero multiple of $\pi$ , infinitely often. The definition of a horizontal asymptote constrains only long-run behavior, so it says the values must close in on 0, not that they must stay on one side of it.

Why is L'Hopital's rule not allowed here?

Its hypotheses require the quotient to be $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , and $\sin x$ approaches neither, since it has no limit as $x \to \infty$ . Applying it anyway gives $\frac{\cos x}{1}$ , which also has no limit, so the rule would suggest the original limit fails to exist. The squeeze theorem shows it does exist and equals 0.

Does the same reasoning work as x approaches negative infinity?

Yes, with one adjustment. For $x < 0$ dividing by $x$ reverses the inequalities, so the bounds swap places and the trap becomes $\frac{1}{x} \le \frac{\sin x}{x} \le -\frac{1}{x}$ . Both bounds still go to 0, so the limit is 0 and $y = 0$ is a horizontal asymptote on both ends.