JEE Main 2018 — Mathematics Calculus

Let, $$\begin{aligned} &L=\lim _{x \rightarrow 0} \frac{(x \tan 2 x-2 x \tan x)}{(1-\cos 2 x)^2}=\lim _{x \rightarrow 0} K \text { (say) } \\ &\Rightarrow K=\frac{x\left[\frac{2 \tan x}{1-(\tan x)^2}\right]-2 x \tan x}{\left(1-\left(1-2 \sin ^2 x\right)\right)^2} \\ &=\frac{2 x \tan x-\left[2 x \tan x-2 x \tan ^3 x\right]}{4 \sin ^4 x \times\left(1-\tan ^2 x\right)} \\ &=\frac{2 x \tan ^3 x}{4 \sin ^4 x \times\left(1-\tan ^2 x\right)} \\ &=\frac{2 x \tan ^3 x}{4 \sin ^4 x \times\left(\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}\right)} \end{aligned}$$​L=x→0lim​(1−cos2x)2(xtan2x−2xtanx)​=x→0lim​K (say) ⇒K=(1−(1−2sin2x))2x[1−(tanx)22tanx​]−2xtanx​=4sin4x×(1−tan2x)2xtanx−[2xtanx−2xtan3x]​=4sin4x×(1−tan2x)2xtan3x​=4sin4x×(cos2xcos2x−sin2x​)2xtan3x​​ $$\begin{aligned} &=\frac{2 x \frac{\sin ^3 x}{\cos ^3 x}}{4 \sin ^4 x \times\left(\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}\right)} \\ & \Rightarrow K=\frac{x}{2 \sin x \times\left(\cos ^2 x-\sin ^2 x\right) \cos x} \\ \therefore L &=\lim _{x \rightarrow 0} \frac{x}{2 \sin x} \times \lim _{x \rightarrow 0} \frac{1}{\cos x\left(\cos ^2 x-\sin ^2 x\right)} \\ &=\lim _{x \rightarrow 0} \frac{x}{2 \sin x} \times \lim _{x \rightarrow 0} \frac{1}{\cos 0\left(\cos ^2 0-\sin ^2 0\right)} \\ &=\frac{1}{2} \end{aligned}$$∴L​=4sin4x×(cos2xcos2x−sin2x​)2xcos3xsin3x​​⇒K=2sinx×(cos2x−sin2x)cosxx​=x→0lim​2sinxx​×x→0lim​cosx(cos2x−sin2x)1​=x→0lim​2sinxx​×x→0lim​cos0(cos20−sin20)1​=21​​