_ x → 0 (1 - 2x) ^ 2 2xtan x - xtan 2x is

$\lim_{x \to 0} \dfrac{{\left(1 - \cos 2x\right)}^{2}}{2xtan x - xtan 2x}$

$= \lim_{x \to 0} \dfrac{{\left(2\sin^{2} x\right)}^{2}}{2x\left(x + \dfrac{{x}^{3}}{3} + \dfrac{2{x}^{5}}{15} + \ldots\right) - x\left(2x + \dfrac{{2}^{3}{x}^{3}}{3} + 2\dfrac{{2}^{5}{x}^{5}}{15} + \ldots\right)}$

$= \lim_{x \to 0} \dfrac{4{\left(x - \dfrac{{x}^{3}}{3 !} + \dfrac{{x}^{5}}{5 !} - \ldots .\right)}^{4}}{{x}^{4}\left(\dfrac{2}{3} - \dfrac{8}{3}\right) + {x}^{6}\left(\dfrac{4}{15} - \dfrac{64}{15}\right)}$

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

$= \lim_{x \to 0} \dfrac{4{\left(1 - \dfrac{{x}^{2}}{3 !} + \dfrac{{x}^{4}}{5 !} - \ldots .\right)}^{4}}{- 2 + {x}^{2}\left(- \dfrac{60}{15}\right) + \ldots}$

$= - 2$