Proof: limit (1-cos x)/x = 0 – Sam Artigliere's Blog

We want to evaluate the expression

$\displaystyle\lim_{x \to 0}\frac{1-\cos x}{x};\quad\text{multiply by}\,1=\frac{1+\cos x}{1+\cos x}\quad\Rightarrow$

$\begin{array}{rcl}\displaystyle\lim_{x \to 0}\frac{1-\cos x}{x}\cdot\frac{1+\cos x}{1+\cos x}&=&\displaystyle\lim_{x \to 0}\,\frac{1-\cos^2 x}{x\cdot(1+\cos x)}\\&=&\displaystyle\lim_{x \to 0}\,\frac{\sin^2 x}{x\cdot(1+\cos x)}\\ &=& \displaystyle\lim_{x \to 0}\,\frac{\sin x \cdot \sin x}{x\cdot(1+\cos x)}\\ &=& \displaystyle\lim_{x \to 0}\frac{\sin x}{x}\cdot\displaystyle\lim_{x \to 0}\frac{\sin x}{1+\cos x}\end{array}$

But

$\displaystyle\lim_{x \to 0}\frac{\sin x}{x}=1$ (see proof)

and

$\displaystyle\lim_{x \to 0}\frac{\sin x}{1+\cos x}=\frac{0}{1+0}=0$

Therefore

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

which is what we sought to find.