Compute specific lim without l'hopital rule

Question

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

Answer 1

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 $1/2$

Answer 2

Hint:

$$(1-\cos z)(1+\cos z)=\sin^2z\implies 1-\cos z=?$$

Use $\lim_{h\to0}\dfrac{\sin h}h=1$