AP Calculus AB and BC

Limit of x cos(1/x) as x Approaches 0

Q: Why can I not just use the product law?

Because it needs BOTH factors to have limits, and \cos(1/x) has none at zero. The squeeze theorem is the tool for a bounded factor with no limit.

The limit of x times the cosine of one over x as x approaches zero is zero. The cosine oscillates forever without settling, but it stays between negative one and one, so multiplying by a vanishing x squeezes the product to zero.

\lim_{x \to 0} x\cos\!\left(\frac{1}{x}\right) = 0

Settled by the squeeze theorem.

Bound the oscillation

The factor $\cos(1/x)$ has NO limit at zero: as $x$ shrinks, $1/x$ races off to infinity and the cosine sweeps the full range from $-1$ to 1 infinitely often. So the product law cannot be used.

-|x| \le x\cos\!\left(\frac{1}{x}\right) \le |x|

Both bounds tend to 0 as $x \to 0$ , so the squeeze theorem forces the trapped function to 0 as well.

Why the bounds must be |x| and not x

For negative $x$ , multiplying the inequality $-1 \le \cos(1/x) \le 1$ by $x$ REVERSES it. Using $|x|$ on both sides sidesteps the sign issue and keeps a single valid pair of bounds for both signs of $x$ .

This is the detail that turns a correct idea into a correct proof, and it is where most written solutions lose a point.

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

Why can I not just use the product law?

Because it needs BOTH factors to have limits, and $\cos(1/x)$ has none at zero. The squeeze theorem is the tool for a bounded factor with no limit.

What about cos(1/x) on its own?

No limit at all. It oscillates across its full range however close you look, which is exactly what the counterexample pages on the squeeze theorem use.