AP Calculus AB and BC

Limit of cos x / x at Infinity Is 0

The limit of cos x over x as x approaches infinity is 0. Cosine never leaves the interval from minus 1 to 1 while the denominator grows without bound, so the quotient is squeezed between minus 1 over x and 1 over x, and both of those go to 0.

lim⁡x→∞cos⁡xx=0\lim_{x \to \infty} \frac{\cos x}{x} = 0

Settled by the squeeze theorem.

Squeezing at infinity

The squeeze theorem is usually taught at a finite point, but it works just as well at infinity. Start from the bound cosine always obeys and divide by a positive xx.

−1x≤cos⁡xx≤1x(x>0)-\frac{1}{x} \le \frac{\cos x}{x} \le \frac{1}{x} \qquad (x > 0)

Both outer functions go to 00 as xx grows, so the function trapped between them has nowhere else to go.

lim⁡x→∞(−1x)=0=lim⁡x→∞1x\lim_{x \to \infty}\left(-\frac{1}{x}\right) = 0 = \lim_{x \to \infty}\frac{1}{x}

Bounded over unbounded is 0

Any bounded numerator divided by something growing without bound goes to 0, whether or not the numerator has a limit of its own. That single rule also settles sin x over x at infinity and arctan x over x.

Why cosine having no limit does not matter

As xx grows, cos⁡x\cos x keeps sweeping between −1-1 and 11 forever, so lim⁡x→∞cos⁡x\lim_{x \to \infty}\cos x does not exist. The quotient law would need that limit, so it is unavailable, and L'Hopital's rule is out too because the form is bounded over infinity rather than ∞∞\frac{\infty}{\infty}.

The graph makes it visible: the oscillations are still there all the way out, but their amplitude is being crushed by the 1x\frac{1}{x} envelope, so the curve threads the horizontal asymptote y=0y = 0 infinitely often.

The mistakes students make

  • Answering that the limit does not exist because cos⁡x\cos x does not. The oscillation continues, but its size is going to 00.
  • Applying L'Hopital's rule. The numerator does not go to infinity, so the form is not one the rule accepts, and differentiating gives −sin⁡x-\sin x over 11, which has no limit at all.
  • Claiming the graph never crosses its asymptote. It crosses y=0y = 0 at every zero of cosine, infinitely many times. Asymptotes describe the trend, not a barrier.

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

What is the limit of cos x / x as x approaches infinity?

It is 00, by the squeeze theorem: −1x≤cos⁡xx≤1x-\frac{1}{x} \le \frac{\cos x}{x} \le \frac{1}{x} for positive xx.

Can I use L'Hopital's rule here?

No. The form is bounded over infinity, not ∞∞\frac{\infty}{\infty}, and differentiating produces −sin⁡x-\sin x, which has no limit.

Does the graph cross y = 0?

Yes, infinitely often, at every zero of cosine. A horizontal asymptote describes long-run behaviour, not a boundary the curve cannot touch.