AP Calculus AB and BC

Limit of sin(x)/x^2 as x Approaches Infinity

The limit of sine x over x squared as x approaches infinity is zero. The numerator never leaves the interval from negative one to one while the denominator grows without bound, so the quotient is squeezed to zero.

lim⁡x→∞sin⁡xx2=0\lim_{x \to \infty} \frac{\sin x}{x^{2}} = 0

Settled by bounded over unbounded.

Bounded over unbounded

−1x2≤sin⁡xx2≤1x2-\frac{1}{x^{2}} \le \frac{\sin x}{x^{2}} \le \frac{1}{x^{2}}

Both bounds tend to 0, so the squeeze theorem gives 0. The oscillation of the numerator never matters, because it is capped and the denominator is not.

The pattern is worth naming: anything bounded, divided by anything growing without bound, tends to 0. It settles a large family of end-behaviour limits with no computation at all.

Why L'Hopital's rule fails here

The form is not indeterminate: the numerator has no limit rather than tending to 0 or infinity, so L'Hopital's hypotheses are not met. Applying it anyway gives cos⁡x2x\frac{\cos x}{2x}, which is no more settled than what you started with.

A rule whose output oscillates as badly as its input is a sign the rule did not apply. The squeeze theorem is the right tool for bounded oscillation.

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

Does this work for cosine too?

Yes, and for any bounded function in the numerator. Only the bound matters, never the oscillation.

What about sin x over x at infinity?

Also 0, by exactly the same argument. Note this is a completely different question from sin⁡xx\frac{\sin x}{x} at ZERO, which is 1.