AP Calculus AB and BC

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

Q: Why multiply by 1 + cos x?

Because (1 - \cos x)(1 + \cos x) = 1 - \cos^2 x = \sin^2 x, and sines are what the special limit knows how to handle. The conjugate is the standard move whenever a 1 \pm \cos or a radical blocks a \frac{0}{0}.

Q: Does L'Hopital's rule work here?

It produces the right number, \lim_{x \to 0} \frac{\sin x}{1} = 0, but differentiating the numerator uses \frac{d}{dx}[\cos x] = -\sin x, and that derivative is proved from this very limit, so the argument is circular. Treat it as a check, not as a justification. The conjugate route is the honest one, it is what Unit 1 expects long before L'Hopital's rule is available, and the same trick clears radicals out of a \frac{0}{0}.

Q: How does this limit give the derivative of sin x?

Expanding \sin(x+h) in the difference quotient leaves \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}. This limit kills the first term and \frac{\sin h}{h} \to 1 keeps the second, which is why the derivative is \cos x.

The limit of (1 minus cos x) divided by x as x approaches 0 is 0. Substitution gives 0/0, which decides nothing. Multiplying by the conjugate 1 plus cos x turns the expression into sin x over x times sin x over 1 plus cos x, and those factors go to 1 and 0. Radians throughout.

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

Settled by the special trigonometric limit.

The conjugate turns this into the sine limit

The numerator $1 - \cos x$ has a conjugate partner $1 + \cos x$ , and their product is $1 - \cos^2 x$ , which the Pythagorean identity rewrites as $\sin^2 x$ . Multiply top and bottom by that partner.

\frac{1-\cos x}{x} \cdot \frac{1+\cos x}{1+\cos x} = \frac{1-\cos^2 x}{x(1+\cos x)} = \frac{\sin^2 x}{x(1+\cos x)}

Split the square so that one sine sits over the $x$ and the other stays with the harmless denominator.

\frac{\sin^2 x}{x(1+\cos x)} = \frac{\sin x}{x} \cdot \frac{\sin x}{1+\cos x}

The first factor is the special limit and goes to 1. The second is continuous at $x = 0$ , so substitution handles it: $\frac{\sin 0}{1 + \cos 0} = \frac{0}{2} = 0$ . Both limits exist, so the product law applies.

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

Why direct substitution fails

At $x = 0$ the numerator is $1 - \cos 0 = 0$ and the denominator is 0, so substitution returns $\frac{0}{0}$ . Landing on that form is not the answer, and it is not evidence that the limit is 0 either, even though 0 turns out to be right.

What makes the answer 0 is a rate mismatch. Near zero $1 - \cos x \approx \frac{x^2}{2}$ , so the numerator vanishes like a square while the denominator vanishes only like $x$ . The quotient behaves like $\frac{x}{2}$ , which collapses to 0.

$x$	$\frac{1-\cos x}{x}$
0.1	0.049958
0.01	0.005000
0.001	0.000500

The values fall by a factor of 10 each time $x$ does, which is the signature of a quotient shrinking like $\frac{x}{2}$ .

Common mistakes

Cancelling part of the numerator. Neither the 1 nor the $\cos x$ is a factor of $1 - \cos x$ , so nothing there cancels with the $x$ underneath.
Confusing it with $\frac{1-\cos x}{x^2}$ , whose limit is $\frac{1}{2}$ . Squaring the denominator matches the numerator's rate exactly and the answer stops being 0.
Multiplying by $1 - \cos x$ instead of $1 + \cos x$ . The conjugate flips the middle sign; repeating the same expression gives $(1-\cos x)^2$ , which simplifies nothing.
Stopping at $\frac{1-\cos^2 x}{x(1+\cos x)}$ . Without rewriting the numerator as $\sin^2 x$ , the special limit never appears.
Assuming the sign matters. $\lim_{x \to 0} \frac{\cos x - 1}{x} = 0$ as well, since a negated zero is still zero. The sign does show up one power down: $\lim_{x \to 0} \frac{\cos x - 1}{x^2} = -\frac{1}{2}$ .

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 multiply by 1 + cos x?

Because $(1 - \cos x)(1 + \cos x) = 1 - \cos^2 x = \sin^2 x$ , and sines are what the special limit knows how to handle. The conjugate is the standard move whenever a $1 \pm \cos$ or a radical blocks a $\frac{0}{0}$ .

Does L'Hopital's rule work here?

It produces the right number, $\lim_{x \to 0} \frac{\sin x}{1} = 0$ , but differentiating the numerator uses $\frac{d}{dx}[\cos x] = -\sin x$ , and that derivative is proved from this very limit, so the argument is circular. Treat it as a check, not as a justification. The conjugate route is the honest one, it is what Unit 1 expects long before L'Hopital's rule is available, and the same trick clears radicals out of a $\frac{0}{0}$ .

How does this limit give the derivative of sin x?

Expanding $\sin(x+h)$ in the difference quotient leaves $\sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}$ . This limit kills the first term and $\frac{\sin h}{h} \to 1$ keeps the second, which is why the derivative is $\cos x$ .