Question

the limit of (1-cosx)/(2(sinx)^2) as x approaches 0

Answer 1

multiply top/bottom by 1+cosx

\[\frac{1-\cos x}{2 \sin^{2} x}*\frac{1+\cos x}{1+\cos x} = \frac{1-\cos^{2} x}{2 \sin^{2} x (1+\cos x)}\]
using pythagorean identity: sin^2 = 1 - cos^2
thus
\[\rightarrow \lim_{x \rightarrow 0} \frac{1}{2(1+\cos x)} = \frac{1}{4}\]

Answer 2

THANK YOU!!