Question

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

Answer 1

$\stackrel{lim}{x\to 0}\frac{xtan2x-2xtanx}{(1-cos2x{)}^2}=\stackrel{lim}{x\to 0}\frac{x.\frac{2tanx}{1-ta{n}^{2}x}-2xtanx}{(si{n}^{2}x{)}^2}L=\frac{2xtanx(\frac{1}{1-ta{n}^{2}x}-1)}{si{n}^{4}x}=\frac{2xtanxxta{n}^{2}x}{si{n}^{2}x(1-ta{n}^{2}x)}L=\stackrel{lim}{x\to 0}\frac{2xta{n}^{3}x}{si{n}^{4}x\left(\frac{co{s}^{2}x-si{n}^{2}x}{co{s}^{2}x}\right)}=\stackrel{lim}{x\to 0}\frac{2xta{n}^{3}x.co{s}^{2}x}{si{n}^{4}x.cos2x}$
$L=\stackrel{lim}{x\to 0}\frac{2xtanx}{cos2x.si{n}^{2}x}$ [Divide N and D is $y{x}^2$]
$L=\stackrel{lim}{x\to 0}\frac{\frac{2xtanx}{cos2x.si{n}^{2}x}}{\frac{cos2x-si{n}^{2}x}{x^2}}=\frac{2\stackrel{lim}{x\to 0}\frac{tanx}{x}}{\left(\stackrel{lim}{x\to 0}cos2x\right)\left(\stackrel{lim}{x\to 0}\frac{si{n}^{2}x}{x^2}\right)}$
$\therefore L=\frac{2\times 1}{1\times 1}$