JEE Main 2016 — Mathematics Calculus

$$\underset{x\rightarrow 0}{\mathrm{lim}}⁡\frac{{(1-\mathrm{cos}⁡2x)}^{2}}{2x\mathrm{tan}⁡x-x\mathrm{tan}⁡2x}$$x→0lim​⁡2xtan⁡x−xtan⁡2x(1−cos⁡2x)2​

$$= \underset{x\rightarrow 0}{\mathrm{lim}}⁡\frac{{(2{\mathrm{sin}}^{2}⁡x)}^{2}}{2x(x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}+\ldots )-x(2x+\frac{{2}^{3} {x}^{3}}{3}+2\frac{{2}^{5} {x}^{5}}{15}+\ldots )}$$=x→0lim​⁡2x(x+3x3​+152x5​+…)−x(2x+323x3​+21525x5​+…)(2sin2⁡x)2​

$$= \underset{x\rightarrow 0}{\mathrm{lim}}⁡\frac{4{(x-\frac{{x}^{3}}{3!}+\frac{{x}^{5}}{5!}-\ldots .)}^{4}}{{x}^{4}(\frac{2}{3}-\frac{8}{3})+{x}^{6}(\frac{4}{15}-\frac{64}{15})}$$=x→0lim​⁡x4(32​−38​)+x6(154​−1564​)4(x−3!x3​+5!x5​−….)4​

(dividing numerator and denominator by ${x}^{4}$x4 )

$$= \underset{x\rightarrow 0}{\mathrm{lim}}⁡\frac{4{(1-\frac{{x}^{2}}{3!}+\frac{{x}^{4}}{5!}-\ldots .)}^{4}}{-2+{x}^{2}(-\frac{60}{15})+\ldots }$$=x→0lim​⁡−2+x2(−1560​)+…4(1−3!x2​+5!x4​−….)4​

$$= -2$$=−2