AP Calculus AB and BC

Limit of x / sin x as x Approaches 0 Is 1

Q: What is the limit of x / sin x as x approaches 0?

It is 1, the reciprocal of \lim_{x\to0}\frac{\sin x}{x} = 1.

Q: Why does it have to be radians?

In degrees \frac{\sin x}{x} \to \frac{\pi}{180}, so this limit would be about 57.3. Every derivative formula for trig functions assumes radians.

Q: What is the limit of x / sin(3x)?

It is \frac{1}{3}. Write it as \frac{1}{3}\cdot\frac{3x}{\sin 3x}, and the second factor tends to 1.

The limit of x over sin x as x approaches 0 is 1. It is the reciprocal of the special limit sin x over x, and since that limit is 1 and 1 is not 0, the reciprocal law applies directly. Like its partner it is only true in radians.

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

Settled by the reciprocal of the special trigonometric limit.

Taking the reciprocal of a known limit

This is the special trigonometric limit turned upside down, so it needs no new work, only a law that has one condition attached.

\lim_{x \to 0}\frac{\sin x}{x} = 1 \implies \lim_{x \to 0}\frac{x}{\sin x} = \frac{1}{1} = 1

The reciprocal law requires the original limit to be nonzero. Here it is $1$ , comfortably away from $0$ , so the step is legal. Had the inner limit been $0$ the reciprocal would have been unbounded instead.

Radians, not degrees

The special limit is 1 only when x is in radians. In degrees sin x over x tends to pi over 180, so x over sin x would tend to 180 over pi, about 57.3. This is why calculus never uses degrees.

Why substitution fails

Substituting $x = 0$ gives $\frac{0}{0}$ , which is indeterminate rather than undefined-and-finished: it means the two parts are racing to zero and the answer depends on their relative speeds.

L'Hopital's rule also works and is quick here, giving $\frac{1}{\cos x} \to 1$ . It is worth knowing the argument is slightly circular if your course proved the derivative of sine using this very limit.

\lim_{x \to 0}\frac{x}{\sin x} \;\overset{\text{H}}{=}\; \lim_{x \to 0}\frac{1}{\cos x} = 1

The mistakes students make

Answering $0$ because the top goes to $0$ . The bottom goes to $0$ at the same rate, which is the whole point.
Using degrees on a calculator to check, which gives about $57.3$ instead of $1$ .
Extending it to $\frac{x}{\sin 3x}$ without adjusting. That limit is $\frac{1}{3}$ , because the inner factor changes the rate.

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 x / sin x as x approaches 0?

It is $1$ , the reciprocal of $\lim_{x\to0}\frac{\sin x}{x} = 1$ .

Why does it have to be radians?

In degrees $\frac{\sin x}{x} \to \frac{\pi}{180}$ , so this limit would be about $57.3$ . Every derivative formula for trig functions assumes radians.

What is the limit of x / sin(3x)?

It is $\frac{1}{3}$ . Write it as $\frac{1}{3}\cdot\frac{3x}{\sin 3x}$ , and the second factor tends to $1$ .