Two limits carry all of trigonometric calculus. We prove the first with the The Squeeze Theorem and a picture of the unit circle, measuring angles in radians.
Key idea
$$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \qquad\text{and}\qquad \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = 0,$$ with $\theta$ in radians. To use them, make the angle inside the sine match the expression in the denominator.
Why it works
Take a small angle $0 < \theta < \frac{\pi}{2}$ on the unit circle. Three regions are nested: the triangle with height $\sin\theta$ (area $\frac{1}{2}\sin\theta$), the circular sector (area $\frac{1}{2}\theta$, since a radian sector of radius 1 has area half its angle), and the larger triangle with height $\tan\theta$ (area $\frac{1}{2}\tan\theta$). So $\sin\theta \le \theta \le \tan\theta$. Divide by $\sin\theta$ and flip: $\cos\theta \le \frac{\sin\theta}{\theta} \le 1$. As $\theta \to 0$, $\cos\theta \to 1$, and the squeeze gives 1. For the second limit, multiply by $\frac{1 + \cos\theta}{1 + \cos\theta}$: $$\frac{1 - \cos\theta}{\theta} = \frac{\sin\theta}{\theta} \cdot \frac{\sin\theta}{1 + \cos\theta} \to 1 \cdot \frac{0}{2} = 0$$.
Worked example: Match the angle
Find $\displaystyle\lim_{x \to 0} \frac{\sin(5x)}{3x}$.
- The angle is $5x$, so we want $5x$ in the bottom.
- Rewrite: $\frac{\sin(5x)}{3x} = \frac{5}{3} \cdot \frac{\sin(5x)}{5x}$.
- As $x \to 0$, $5x \to 0$, so $\frac{\sin(5x)}{5x} \to 1$.
Answer: $\frac{5}{3}$
Worked example: Tangent
Find $\displaystyle\lim_{x \to 0} \frac{\tan(2x)}{x}$.
- $\tan(2x) = \frac{\sin(2x)}{\cos(2x)}$, so the expression is $2 \cdot \frac{\sin(2x)}{2x} \cdot \frac{1}{\cos(2x)}$.
- The factors tend to $2 \cdot 1 \cdot 1$.
Answer: $2$
Worked example: A squared bottom
Find $\displaystyle\lim_{x \to 0} \frac{1 - \cos(4x)}{x^2}$.
- Multiply by $1 + \cos(4x)$: the top becomes $1 - \cos^2(4x) = \sin^2(4x)$.
- So the expression is $\left(\frac{\sin(4x)}{x}\right)^2 \cdot \frac{1}{1 + \cos(4x)}$.
- $\frac{\sin(4x)}{x} = 4 \cdot \frac{\sin(4x)}{4x} \to 4$, and $1 + \cos(4x) \to 2$.
- The limit is $\frac{16}{2} = 8$.
Answer: $8$
| $x$ (radians) | $0.5$ | $0.1$ | $0.01$ |
|---|---|---|---|
| $\frac{\sin x}{x}$ | 0.9589 | 0.9983 | 0.99998 |
| $\frac{1 - \cos x}{x}$ | 0.2448 | 0.0500 | 0.0050 |
Common mistake
Two traps. First, "canceling the $x$" in $\frac{\sin x}{x}$ to get $\sin$: the $x$ inside $\sin(x)$ is an input, not a factor. Second, working in degrees. If $x$ is in degrees, $\frac{\sin x}{x} \to \frac{\pi}{180} \approx 0.01745$, not 1. Check with a calculator in radian mode: $\frac{\sin(0.001)}{0.001} = 0.99999983\ldots$
These two limits are exactly the tools for differentiating sine and cosine: the slope of $\sin x$ at 0 is $\lim_{h \to 0} \frac{\sin h}{h} = 1$, and the second limit takes care of the cosine.