AP Calculus AB and BC

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

Q: Does L'Hopital's rule confirm the answer?

It does, in two passes. First \frac{2\sin 2x}{2x} = \frac{\sin 2x}{x}, still \frac{0}{0}, then \frac{2\cos 2x}{1} \to 2. Unit 1 expects the identity route, since L'Hopital is a Unit 4 tool, but it is a fair check.

Q: What is the limit of (1 - cos ax)/x^2 for a general a?

It is \frac{a^{2}}{2}. Write it as a^{2} \cdot \frac{1-\cos ax}{(ax)^{2}} and the fraction matches the standard limit \frac{1}{2} because ax \to 0. So a = 1 gives \frac{1}{2}, a = 2 gives 2, and a = 3 gives \frac{9}{2}.

Q: Why is (1 - cos 2x)/x zero while this one is not?

Degree counting. The numerator behaves like 2x^{2}, so dividing by x leaves something like 2x, which goes to 0, while dividing by x^{2} leaves the constant 2. One power of x decides between a limit of 0 and a limit of 2.

The limit of (1 - cos 2x)/x^2 as x approaches 0 is 2. Direct substitution gives 0 over 0. The double-angle identity turns the numerator into 2 sin squared x, so the expression is twice the square of sin x over x. The inner coefficient enters squared, which is why the answer is 2 and not one half.

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

Settled by the special trigonometric limit.

The identity that removes the cosine

The double-angle formula $\cos 2x = 1 - 2\sin^{2}x$ rearranges into exactly the numerator, and it converts a subtraction that vanishes into a product that can be split.

1-\cos 2x = 2\sin^{2}x

Dividing by $x^{2}$ now produces a perfect square of the one trig limit everyone memorises.

\frac{1-\cos 2x}{x^{2}} = \frac{2\sin^{2}x}{x^{2}} = 2\left(\frac{\sin x}{x}\right)^{2}

Both factors now have limits, so the product law applies and the special limit finishes it.

\lim_{x \to 0}2\left(\frac{\sin x}{x}\right)^{2} = 2 \cdot 1^{2} = 2

Rescaling the standard result

If you already know $\lim_{u \to 0}\frac{1-\cos u}{u^{2}} = \frac{1}{2}$ , substitute $u = 2x$ and keep the bookkeeping honest: $\frac{1-\cos 2x}{x^{2}} = 4 \cdot \frac{1-\cos 2x}{(2x)^{2}}$ , which goes to $4 \cdot \frac{1}{2} = 2$ . The factor is 4, not 2, because the denominator is squared.

Why direct substitution fails

At $x = 0$ the cosine equals 1, so the numerator collapses along with the denominator.

\frac{1-\cos 0}{0^{2}} = \frac{1-1}{0} = \frac{0}{0}

The form says only that both parts vanish, not how fast. Here the numerator vanishes to second order, because $\cos u \approx 1 - \frac{u^{2}}{2}$ makes $1 - \cos 2x \approx 2x^{2}$ , and the denominator is second order too. Matched orders give a finite nonzero answer.

$x$	$\frac{1-\cos 2x}{x^{2}}$
0.1	1.993342
0.01	1.999933
0.001	1.999999

Values approach from below, and the function is even, so the left side gives the same numbers. A calculator in degree mode will not reproduce this table, since the special trig limit and the small-angle expansion both assume radians.

The mistakes students make

Answering $\frac{1}{2}$ , straight from the memorised $\frac{1-\cos x}{x^{2}} \to \frac{1}{2}$ , with no adjustment for the inner 2.
Answering 1, by pulling the inner coefficient out linearly as $2 \cdot \frac{1}{2}$ . The coefficient enters squared against an $x^{2}$ denominator, so the general result is $\lim_{x \to 0}\frac{1-\cos ax}{x^{2}} = \frac{a^{2}}{2}$ .
Misquoting the identity as $1-\cos 2x = 2\cos^{2}x$ . The correct forms are $\cos 2x = 1-2\sin^{2}x$ and $\cos 2x = 2\cos^{2}x-1$ ; both give $1-\cos 2x = 2\sin^{2}x$ .
Splitting into $\frac{1}{x^{2}} - \frac{\cos 2x}{x^{2}}$ and taking limits term by term. Each piece diverges, and the difference law needs both limits to exist.
Cancelling the $x$ in $\cos 2x$ against the denominator. The $x$ there is an argument, not a factor.

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 L'Hopital's rule confirm the answer?

It does, in two passes. First $\frac{2\sin 2x}{2x} = \frac{\sin 2x}{x}$ , still $\frac{0}{0}$ , then $\frac{2\cos 2x}{1} \to 2$ . Unit 1 expects the identity route, since L'Hopital is a Unit 4 tool, but it is a fair check.

What is the limit of (1 - cos ax)/x^2 for a general a?

It is $\frac{a^{2}}{2}$ . Write it as $a^{2} \cdot \frac{1-\cos ax}{(ax)^{2}}$ and the fraction matches the standard limit $\frac{1}{2}$ because $ax \to 0$ . So $a = 1$ gives $\frac{1}{2}$ , $a = 2$ gives 2, and $a = 3$ gives $\frac{9}{2}$ .

Why is (1 - cos 2x)/x zero while this one is not?

Degree counting. The numerator behaves like $2x^{2}$ , so dividing by $x$ leaves something like $2x$ , which goes to 0, while dividing by $x^{2}$ leaves the constant 2. One power of $x$ decides between a limit of 0 and a limit of 2.