Evaluate the limit limx→0cos(ax)−1sin2(bx)\lim_{x \to 0} \frac{\cos(ax) - 1}{\sin^2(bx)}limx→0sin2(bx)cos(ax)−1 where a,b≠0a, b \neq 0a,b=0.
−a22b2-\frac{a^2}{2b^2}−2b2a2
−a2b2-\frac{a^2}{b^2}−b2a2
ab\frac{a}{b}ba
a22b2\frac{a^2}{2b^2}2b2a2