Question

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

Answer 1

$\underset{x\to 0}{lim}\frac{xtan2x-2xtanx}{(1-cos2x{)}^2}$

$1-cos2x=2si{n}^{2}x$

$\underset{x\to 0}{lim}\frac{x(tan2x-2tanx)}{\left(2si{n}^{2}x\right)}$

$[\Rightarrow tanx=\frac{2tanx}{1-ta{n}^{2}x}]$

$\underset{x\to 0}{lim}\frac{x(\frac{2tanx}{1-ta{n}^{2}x}-2tanx)}{(2si{n}^{2}x{)}^2}$

$\underset{x\to 0}{lim}\frac{x(2tanx-2tanx+2ta{n}^{3}x)}{\left(2si{n}^{2}x{)}^2\right(1-ta{n}^{2}x)}$

$\underset{x\to 0}{lim}\frac{x2ta{n}^{3}x}{(1-ta{n}^{2}x)\left(4si{n}^{2}x\right)}$

$\underset{x\to 0}{lim}\frac{x2si{n}^{2}x}{ta{n}^{2}x(1-ta{n}^{2}x)\left(1/2si{n}^{2}x\right)}$

divide numerator and denominator by x

$\underset{x\to 0}{lim}\frac{1}{2co{s}^{3}x(1-ta{n}^{2}x)\left(\frac{sinx}{x}\right)}$

$\underset{x\to 0}{lim}\frac{sinx}{x}=1,cos\left(0\right)=1,tan0=0$

Applying limit $x\to 0$

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

$=1/2$