Use l' Hopital's Rule to find the limit. Show your work.
Lim cosx-1/2x^2 is 0/0 form
x->0
In applying L'Hopitals rule, you need to differentiate both numerator and denominator of the limit. Thus we see that
$\displaystyle \lim_{x\to0}\frac{\cos\!\left(x\right)-1}{2x^2}=\lim_{x\to0}\frac{-\sin\!\left(x\right)}{4x}$
Again, this limit has the indeterminate form $\displaystyle \frac{0}{0}$
So, I leave it for you to apply the rule one more time.
Does this make sense?