Special Trigonometric Limits
Prove and apply lim(x→0) sin(x)/x = 1 and lim(x→0) (1−cos(x))/x = 0 geometrically. Use these foundational results to derive the derivatives of sine and cosine later in the course.
A lesson of Calculus, in Limits and Continuity, about 10 minutes.
By the end of this lesson, you will be able to prove $\lim_{x\to 0}\sin x/x = 1$ using the Squeeze Theorem, derive related limits, and apply them to evaluate complex trig limits.
The Two Fundamental Trig Limits
$$\lim_{x \to 0}\frac{\sin x}{x} = 1 \qquad \lim_{x \to 0}\frac{1 - \cos x}{x} = 0$$
These are indeterminate ($0/0$) at $x = 0$ and cannot be resolved by factoring. They are proved using the Squeeze Theorem and geometric arguments.
Proof of $\lim_{x \to 0}\frac{\sin x}{x} = 1$
Consider the unit circle for $0 < x < \pi/2$.
Area comparison:
$\text{Area of triangle } OAP < \text{Area of sector } OAP < \text{Area of triangle } OAQ$
$$\frac{1}{2}\sin x < \frac{1}{2}x < \frac{1}{2}\tan x$$
Dividing by $\frac{1}{2}\sin x$ (positive):
$$1 < \frac{x}{\sin x} < \frac{1}{\cos x}$$
Taking reciprocals (reversing inequalities):
$$\cos x < \frac{\sin x}{x} < 1$$
As $x \to 0^+$: $\cos x \to 1$ and $1 \to 1$, so by Squeeze Theorem: $\dfrac{\sin x}{x} \to 1$.
For $x \to 0^-$: use the fact that $\sin x/x$ is an even function. QED. ∎
Deriving $\lim_{x \to 0}\frac{1-\cos x}{x} = 0$
Multiply by $\dfrac{1+\cos x}{1+\cos x}$:
$$\frac{1-\cos x}{x} \cdot \frac{1+\cos x}{1+\cos x} = \frac{\sin^2 x}{x(1+\cos x)} = \frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x}$$
As $x \to 0$: $\dfrac{\sin x}{x} \to 1$ and $\dfrac{\sin x}{1+\cos x} \to 0/2 = 0$.
Product: $1 \cdot 0 = 0$. ✓
Applications
$\lim_{x \to 0}\dfrac{\sin(5x)}{3x}$:
$= \dfrac{5}{3}\lim_{x \to 0}\dfrac{\sin(5x)}{5x} = \dfrac{5}{3}(1) = \dfrac{5}{3}$
$\lim_{x \to 0}\dfrac{\tan x}{x}$:
$= \lim_{x \to 0}\dfrac{\sin x}{x\cos x} = \dfrac{1}{1} = 1$
$\lim_{x \to 0}\dfrac{1-\cos(2x)}{x^2}$:
$= \lim_{x \to 0}\dfrac{1-\cos(2x)}{(2x)^2/4} \cdot \dfrac{1}{1} = 4\lim_{u \to 0}\dfrac{1-\cos u}{u^2}$
Since $\lim_{u \to 0}\dfrac{1-\cos u}{u^2} = \dfrac{1}{2}$, this evaluates to $4 \cdot \dfrac{1}{2} = 2$.
A cleaner route uses the identity $1 - \cos(2x) = 2\sin^2 x$: $\dfrac{1-\cos(2x)}{x^2} = 2\left(\dfrac{\sin x}{x}\right)^2 \to 2$ as $x \to 0$.
Example — Composite Argument
$\lim_{x \to 0}\dfrac{\sin(3x)}{\sin(5x)}$:
$= \lim_{x \to 0}\dfrac{\sin(3x)}{3x}\cdot\dfrac{5x}{\sin(5x)}\cdot\dfrac{3x}{5x} = 1\cdot 1\cdot\dfrac{3}{5} = \dfrac{3}{5}$
Connection to the Derivative of Sine
$$\frac{d}{dx}\sin x\bigg|{x=0} = \lim{h \to 0}\frac{\sin(0+h)-\sin 0}{h} = \lim_{h \to 0}\frac{\sin h}{h} = 1$$