1.limx→0sin2x⋅cot22xx⋅cot4x\displaystyle\lim_{x\to0}\frac{\sin^2x\cdot\cot^2 2x}{x\cdot\cot4x}x→0limx⋅cot4xsin2x⋅cot22x is equal toa.000b.111c.444d.14\frac1441Login to continueOnly logged in users canattempt or see the solution.