first we try to evalaute
doesn't work
so we use l'hopital's rule
because we get 0/0 or an inditerminate form when evaluting the limit
so
for f(x)/g(x), we can do f'(x)/g'(x) and evaluate again
takeing the derivitive of the top and bottom sepearely, we get
[tex]\frac{sin(x)}{2sin(x)cos(x)}[/tex]
evaluating, we get 0/0
another inditerminate
take derivitive of top and bottom again
use chain rule
we get
[tex]\frac{cos(x)}{2(cos^2(x)-sin^2(x))}[/tex]
evaluating, we get 1/(2(1-0))=1/(2)=1/2
[tex] \lim_{x \to 0}\frac{1-cos}{sin^2(x)}=\frac{1}{2}[/tex]
