Using $\epsilon-\delta$ language to show $\lim_{x\rightarrow0}\frac{1-\cos(x)}{x}=0$.

Question

I've permitted myself to use $\displaystyle\lim_{x\rightarrow0}\frac{\sin(x)}{x}=1$, as I've proven this.

Here's some manipulation I've tried, but I can't quite get the $1+\cos(x)$ terms in the denominators bounded, so I don't think this approach will be fruitful:

$$|\frac{1-\cos(x)}{x}|=|\frac{1-\cos^2(x)}{x(1+\cos(x))}|=|\frac{\sin^2(x)}{x(1+\cos(x))}|=|\frac{\sin(x)}{x}\cdot\frac{\sin(x)}{1+\cos(x)}|=|\frac{\sin(x)}{x}\cdot\frac{\sin(x)}{1+\cos(x)}|=|\frac{\sin(x)-x+x}{x}\cdot\frac{\sin(x)}{1+\cos(x)}|=|(\frac{\sin(x)-x}{x}+1)\cdot(\frac{\sin(x)}{1+\cos(x)})|=|\frac{\sin(x)-x}{x}\cdot \frac{\sin(x)}{1+\cos(x)}+\frac{\sin(x)}{1+\cos(x)}|$$

Answer 1

Using inequality $$|\sin\alpha|\leq|\alpha|$$ then $$\left|\frac{1-\cos(x)}{x}\right|=\left|\dfrac{2\sin^2\frac{x}{2}}{x}\right|\leq\left|\dfrac{2\left(\frac{x}{2}\right)^2}{x}\right|=\left|\frac{x}{2}\right|$$