Proof of Trig Limits — Question 1

Question 1 - Solution

We prove this limit using geometric inequalities.

Consider a unit circle centered at the origin.

Let $0<x<\pi/2$ be measured in radians, and consider the angle $x$.

The area of the sector with angle $x$ is

$$\text{Area of sector}=\frac{1}{2}x.$$

The area of the triangle formed by the radius and the angle $x$ is

$$\text{Area of triangle}=\frac{1}{2}\sin x.$$

The area of the triangle formed by the tangent line is

$$\text{Area of outer triangle}=\frac{1}{2}\tan x.$$

Thus, the areas satisfy

$$\frac{1}{2}\sin x < \frac{1}{2}x < \frac{1}{2}\tan x.$$

Multiply through by $2$:

$$\sin x < x < \tan x.$$

Divide all parts by $\sin x$ (which is positive for $0<x<\pi/2$):

$$1 < \frac{x}{\sin x} < \frac{1}{\cos x}.$$

Taking reciprocals reverses the inequalities:

$$\cos x < \frac{\sin x}{x} < 1.$$

Now take the limit as $x\to 0^+$.

Since

$$\lim_{x\to 0}\cos x=1,$$

the Squeeze Theorem gives

$$\lim_{x\to 0^+}\frac{\sin x}{x}=1.$$

A similar argument applies for $x\to 0^-$, so the two-sided limit exists.

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