Squeeze Theorem sinx over x

WTS

${\lim }_{x\to 0}\frac{\sin x}{x}$

Proof with Squeeze Theorem and Geometry

We make 3 triangles, the middle triangle is f(x) which we will squeeze between h(x) and g(x) We want $g(x)\le f(x)\le h(x)$ So

• $g(x)=\frac{1}{2}(1)(\tan x)$
• $f(x)=\frac{1}{2}(1)(x)$
• $h(x)=\frac{1}{2}(1)(\sin x)$ Then $\sin x\le x\le \tan x$ (factored out $\frac{1}{2}$) Then $1\le \frac{x}{\sin x}\le \frac{1}{\cos x}$ Then $1\ge \frac{\sin x}{x}\ge \cos x$ Then ${\lim }_{x\to 0}\cos x=1$ Since $1\le \frac{\sin x}{x}\le 1$ Using squeeze theorem, $\frac{\sin x}{x}$