Evaluating $\lim\limits_{x\to0}\frac{1-\cos(x)}{x}$ $$\lim_{x\to0}\frac{1-\cos(x)}{x}$$
Could someone help me with this trigonometric limit? I am trying to evaluate it without l'Hôpital's rule and derivation.

There is also a fairly direct method based on trig identities and the limit $ \ \lim_{\theta \rightarrow 0} \ \tan \frac{\theta}{2} \ = \ 0 , $ which I discuss in the first half of my post here.  
[In brief, 
$$ \lim_{\theta \rightarrow 0} \ \tan \frac{\theta}{2} \ = \ 0 \ = \ \lim_{\theta \rightarrow 0} \ \frac{1 - \cos \theta}{\sin \theta}  \ = \ \lim_{\theta \rightarrow 0} \ \frac{1 - \cos \theta}{\theta}  \ \cdot \ \frac{\theta}{\sin \theta} $$
$$= \ \lim_{\theta \rightarrow 0} \ \frac{1 - \cos \theta}{\theta}  \ \cdot \ \frac{1}{\lim_{\theta \rightarrow 0} \ \frac{\sin \theta}{\theta} } \ = \ \lim_{\theta \rightarrow 0} \ \frac{1 - \cos \theta}{\theta}  \ \cdot \ \frac{1}{1} \ \Rightarrow  \ \lim_{\theta \rightarrow 0} \ \frac{1 - \cos \theta}{\theta} \ = \ 0 \ . ] $$
