Compute specific lim without l'hopital rule how can i compute this limit without L'hopital rule:

$\lim \limits _{z\to 0} \frac{1-cos z}{z^2}$ 

i believe the answer is 1/2 by the rule, 

thanks
Remember that

$$1-\cos{z}=2\sin^2{z\over 2}$$

And so the limit we are looking for is

$$\lim_{z\to 0}{1\over 2}{\sin^2{z\over 2}\over \left({z\over 2}\right)^2}=\lim_{z\to 0}{1\over 2}\left({\sin{z\over 2}\over {z\over 2}}\right)^2$$

Now ${\sin{X}\over X}\to 1$ and so the limit is indeed $1/2$
Very easy, just use the fact that near $0$, the numerator $1 -\cos z$ is asymptotically equivalent to $\large{{1\over2}z^2}$ so we have:

$\Large{1 -\cos z \over z^2}$ $\sim $ $\Large{{1\over2}z^2 \over z^2}$ $=$ $\Large{1\over2}$ as $x \to 0$