Mathematics - Limits Question with Solution | TestHub

$\underset{x\to 0}{\lim }\frac{{\left(1-\cos 2x\right)}^2}{2x\tan x-x\tan 2x}$

$= \underset{x\to 0}{\lim }\frac{{\left(2{\sin }^{2}x\right)}^2}{2x\left(x+\frac{x^3}{3}+\frac{2x^5}{15}+\dots \right)-x\left(2x+\frac{2^3 x^3}{3}+2\frac{2^5 x^5}{15}+\dots \right)}$

$= \underset{x\to 0}{\lim }\frac{4{\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots .\right)}^4}{x^4\left(\frac{2}{3}-\frac{8}{3}\right)+x^6\left(\frac{4}{15}-\frac{64}{15}\right)}$

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

$= \underset{x\to 0}{\lim }\frac{4{\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\dots .\right)}^4}{-2+x^2\left(-\frac{60}{15}\right)+\dots }$

$= -2$