AP Calculus AB and BC

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

Q: Why is the answer 1/2 rather than 0?

Because both parts vanish at the same order. Near zero 1 - \cos x behaves like \frac{x^2}{2}, so dividing by x^2 leaves roughly \frac{1}{2} rather than something that shrinks. Compare \frac{1-\cos x}{x}, where the numerator vanishes faster than the denominator and the limit really is 0.

Q: Can I evaluate it without L'Hopital's rule?

Yes. Multiply by \frac{1+\cos x}{1+\cos x} to get \left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1+\cos x}, then use the special limit and continuity: 1 \cdot \frac{1}{2} = \frac{1}{2}. That version is legal in Unit 1.

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

It is -\frac{1}{2}. The numerator is the negative of the one here, so the limit flips sign. Reversing the order of the two terms is a common sign slip on free response.

The limit of (1 minus cos x) divided by x squared as x approaches 0 is one half. Substitution gives 0/0, and so does the quotient after one round of L'Hopital, so the rule has to be applied twice. The answer is not 0: the numerator shrinks like x squared over 2, matching the denominator's rate.

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

Settled by L'Hopital's rule applied twice.

Two applications, with the form checked each time

Numerator and denominator are differentiable near 0 and both go to 0, so the hypotheses hold. Differentiate the top and the bottom separately, which is not the quotient rule.

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

Substituting into the new quotient gives $\frac{0}{0}$ again, so the hypotheses hold a second time and the rule fires again.

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

That last quotient is continuous at 0, so substitution finishes it. The second round is also avoidable: $\frac{\sin x}{2x} = \frac{1}{2} \cdot \frac{\sin x}{x} \to \frac{1}{2}$ by the special limit.

The conjugate route, no L'Hopital needed

This limit appears in Unit 1, before L'Hopital shows up in Topic 4.7, so the algebraic version is worth having: $\frac{1-\cos x}{x^2} \cdot \frac{1+\cos x}{1+\cos x} = \frac{\sin^2 x}{x^2(1+\cos x)} = \left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1+\cos x}$ , which tends to $1^2 \cdot \frac{1}{2}$ .

Why substitution fails, and why one round is not enough

Substituting $x = 0$ gives $\frac{1 - \cos 0}{0^2} = \frac{0}{0}$ . The trap sits one step later: after differentiating once you have $\frac{\sin x}{2x}$ , which still evaluates to $\frac{0}{0}$ , and a student who reads that as 0 reports the wrong limit.

The size of the answer comes from a rate match. The Maclaurin series $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$ gives $1 - \cos x = \frac{x^2}{2} - \frac{x^4}{24} + \cdots$ , so dividing by $x^2$ leaves $\frac{1}{2} - \frac{x^2}{24} + \cdots$ . Numerator and denominator vanish at exactly the same order, and the leading coefficients decide the value.

$x$	$\frac{1-\cos x}{x^2}$
0.1	0.499583
0.01	0.499996
0.001	0.500000

Common mistakes

Answering 0 because the numerator goes to 0. The denominator goes to 0 as well, and $\frac{0}{0}$ has no default value.
Stopping after one application. $\frac{\sin x}{2x}$ is still indeterminate at $x = 0$ ; it needs the special limit or a second round of the rule.
Using the quotient rule. L'Hopital differentiates the numerator and the denominator independently, so the derivative of $\frac{1-\cos x}{x^2}$ never enters the work.
Applying the rule a third time. Once the quotient is $\frac{\cos x}{2}$ , substitution gives $\frac{1}{2}$ and the form is no longer indeterminate; differentiating again would produce $\frac{-\sin x}{0}$ , which is meaningless.
Borrowing the answer for $\frac{1-\cos x}{x}$ . Same numerator, one power less underneath, and the limit drops to 0.

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 is the answer 1/2 rather than 0?

Because both parts vanish at the same order. Near zero $1 - \cos x$ behaves like $\frac{x^2}{2}$ , so dividing by $x^2$ leaves roughly $\frac{1}{2}$ rather than something that shrinks. Compare $\frac{1-\cos x}{x}$ , where the numerator vanishes faster than the denominator and the limit really is 0.

Can I evaluate it without L'Hopital's rule?

Yes. Multiply by $\frac{1+\cos x}{1+\cos x}$ to get $\left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1+\cos x}$ , then use the special limit and continuity: $1 \cdot \frac{1}{2} = \frac{1}{2}$ . That version is legal in Unit 1.

What is the limit of (cos x - 1)/x^2?

It is $-\frac{1}{2}$ . The numerator is the negative of the one here, so the limit flips sign. Reversing the order of the two terms is a common sign slip on free response.