The Squeeze Theorem and Special Trigonometric Limits

Special Trigonometric Limits

By the end of this lesson, you will be able to evaluate limits involving trigonometric functions by applying the two fundamental trig limits.

The Two Fundamental Limits

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

These are proved using the Squeeze Theorem (and geometric area arguments). They must be memorized.

Why do they matter? They are the foundation for the derivative of $\sin x$ and appear throughout calculus and physics.

Applying the First Limit

Key: the argument of sin and the denominator must match.

$\lim_{x\to 0}\frac{\sin(3x)}{x}$

Multiply and divide by 3: $= 3\lim_{x\to 0}\frac{\sin(3x)}{3x} = 3(1) = 3$

$\lim_{x\to 0}\frac{\sin(5x)}{\sin(3x)} = \lim_{x\to 0}\frac{\sin(5x)}{5x} \cdot \frac{3x}{\sin(3x)} \cdot \frac{5}{3} = 1 \cdot 1 \cdot \frac{5}{3} = \frac{5}{3}$

$\lim_{\theta\to 0}\frac{\tan\theta}{\theta} = \lim_{\theta\to 0}\frac{\sin\theta/\cos\theta}{\theta} = \lim_{\theta\to 0}\frac{\sin\theta}{\theta} \cdot \frac{1}{\cos\theta} = 1 \cdot 1 = 1$

Using the Second Limit

$\lim_{x\to 0}\frac{1-\cos(2x)}{x} = 2\lim_{x\to 0}\frac{1-\cos(2x)}{2x} = 2(0) = 0$

Connection to the Derivative of sin x

The derivative of $\sin x$ is:

$\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h} = \lim_{h\to 0}\frac{\sin x\cos h + \cos x\sin h - \sin x}{h}$

$= \sin x\underbrace{\lim_{h\to 0}\frac{\cos h - 1}{h}}{=0} + \cos x\underbrace{\lim{h\to 0}\frac{\sin h}{h}}_{=1} = \cos x$

Summary

$\lim_{x\to 0}\frac{\sin x}{x} = 1$ (must be memorized).

$\lim_{x\to 0}\frac{1-\cos x}{x} = 0$ (must be memorized).

Technique: rewrite so the argument of sin matches the denominator, then apply the limits.

$\lim\frac{\sin(kx)}{x} = k$; $\lim\frac{\sin(kx)}{kx} = 1$.