AP Calculus AB and BC

Limit of (cos x - 1) / sin x at 0 Is 0

Q: Why do the two zeros not cancel to 1?

They vanish at different rates: \cos x - 1 behaves like -\frac{x^{2}}{2} while \sin x behaves like x, so the quotient behaves like -\frac{x}{2}.

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

It is \frac{1}{2}, a different standard limit with a squared denominator.

The limit of cos x minus 1, over sin x, as x approaches 0 is 0. Divide the numerator and denominator by x: the top becomes the standard limit that goes to 0 and the bottom becomes the one that goes to 1, so the quotient is 0 over 1.

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

Settled by dividing through by x to expose the standard limits.

Dividing through by x

Both parts vanish at $0$ , so the form is $\frac{0}{0}$ . Dividing top and bottom by $x$ costs nothing and turns each part into a limit you already know.

\frac{\cos x - 1}{\sin x} = \frac{\frac{\cos x - 1}{x}}{\frac{\sin x}{x}}

The numerator tends to $0$ and the denominator tends to $1$ , and because the denominator's limit is nonzero the quotient law applies.

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

Both parts vanish, but not at the same rate

Near the origin cos x minus 1 behaves like minus x squared over 2, while sin x behaves like x. A quadratic over a linear goes to 0, which is the real reason the answer is 0.

The half-angle route

Multiplying by the conjugate gives a second derivation and shows the sign clearly.

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

That last expression is continuous at $0$ , so substitution finishes it: $\frac{-0}{2} = 0$ . It also shows the function approaches $0$ from below for small positive $x$ .

The mistakes students make

Reporting that the limit does not exist because both parts are $0$ . The rates differ, and that is what decides the answer.
Flipping the sign. The numerator $\cos x - 1$ is NEGATIVE near $0$ , so the function is negative on the right side.
Confusing it with $\frac{1 - \cos x}{x^{2}}$ , which is $\frac{1}{2}$ . Different denominator, different answer.

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 limit of (cos x - 1) / sin x as x approaches 0?

It is $0$ .

Why do the two zeros not cancel to 1?

They vanish at different rates: $\cos x - 1$ behaves like $-\frac{x^{2}}{2}$ while $\sin x$ behaves like $x$ , so the quotient behaves like $-\frac{x}{2}$ .

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

It is $\frac{1}{2}$ , a different standard limit with a squared denominator.