AP Calculus AB and BC

Limit of (1-cos 3x)/x^2 as x Approaches 0

Q: What is the general formula?

\lim_{x \to 0}\frac{1-\cos kx}{x^{2}} = \frac{k^{2}}{2}. Derive it by matching the denominator to (kx)^{2} and pulling out k^{2}.

Q: Could L'Hopital's rule do this?

Yes, applied twice, since 0/0 persists after the first differentiation. The matching method is faster and does not need the form re-checked between steps.

The limit of one minus cosine of 3x over x squared as x approaches zero is nine halves. The standard cosine limit needs the denominator to match the inner angle, and forcing that match leaves a factor of nine behind.

\lim_{x \to 0} \frac{1-\cos 3x}{x^{2}} = \frac{9}{2}

Settled by matching the inner angle to the denominator.

Force the match

The standard limit is $\lim_{u \to 0}\frac{1-\cos u}{u^{2}} = \frac{1}{2}$ , and it only applies when the denominator is the SQUARE of whatever sits inside the cosine. Here the inside is $3x$ , so the denominator wants $(3x)^{2} = 9x^{2}$ .

\frac{1-\cos 3x}{x^{2}} = 9 \cdot \frac{1-\cos 3x}{9x^{2}} = 9 \cdot \frac{1-\cos 3x}{(3x)^{2}}

Now the standard limit applies to the second factor and returns $\frac{1}{2}$ , so the whole thing tends to $9 \cdot \frac{1}{2} = \frac{9}{2}$ .

Why the factor is nine and not three

The denominator is squared, so the correction is squared too. With $\frac{1-\cos kx}{x^{2}}$ the answer is always $\frac{k^{2}}{2}$ , which is why $k = 3$ gives $9/2$ rather than $3/2$ .

Compare the sine version: $\frac{\sin kx}{x} \to k$ , with no squaring, because that denominator is only first power. Mixing up the two is the standard error here.

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 general formula?

$\lim_{x \to 0}\frac{1-\cos kx}{x^{2}} = \frac{k^{2}}{2}$ . Derive it by matching the denominator to $(kx)^{2}$ and pulling out $k^{2}$ .

Could L'Hopital's rule do this?

Yes, applied twice, since $0/0$ persists after the first differentiation. The matching method is faster and does not need the form re-checked between steps.