1.The value of limx→01+sinx−cosx+ln(1−x)xtan2x\displaystyle\lim_{x\to0}\frac{1+\sin x-\cos x+\ln(1-x)}{x\tan^2x}x→0limxtan2x1+sinx−cosx+ln(1−x) isa.−1/2-1/2−1/2b.−1/3-1/3−1/3c.1/21/21/2d.1/41/41/4Login to continueOnly logged in users canattempt or see the solution.
