I need help in the following problem:
$$\lim_{x\to0} \frac{x - x \cos x}{\sin^2 2 x}.$$ L'Hopital has not yet given in my Calculus I class at college, so there has to be a way to solve it without L'Hopital, but I can't find it.
My proof:
$$\lim_{x\to0} \frac{x - x \cos x}{\sin^2 2 x} = \frac{x.(1-cosx)(1+cosx)}{sin^2(2x).(1+cosx)}=\frac{x.sin^2x}{(1+cosx).sin^2(2x)}$$ $$lim_{x\to0}\frac {x - x \cos x}{\sin^2 2 x}= \frac{x}{1+cosx}. [{\frac{sinx}{sin(2x)}]^2}$$ $$\lim_{x\to0}\frac {x - x \cos x}{\sin^2 2 x}=\frac{0}{2}.\frac{1}{4}=0$$