Question

Evaluate Limx------>0 tan x- sin x/ x(1-cos2x)

Answer 1

Answer:

We have to find,

$\lim_{x \to 0} \frac{tan x-sin x}{x(1-cos 2 x)}\\\\ \lim_{x \to 0} \frac{tanx(1-cosx)}{x(1-cos 2 x)}\\\\ \lim_{x \to 0}\frac{tanx}{x} \times  \lim_{x \to 0}\frac{1-cosx}{1-cos 2x}\\\\ 1 \times\lim_{x \to 0} \frac{sin^2(\frac{x}{2})}{sin^2x}\\\\ \lim_{x \to 0} \frac{\frac{4\times sin^2(\frac{x}{2})^2}{\frac{x^2}{4}}}{\frac{sin^2x}{x^2}}\\\\=\frac{1}{4}$

${\text{because,}\lim_{x \to 0 }\frac{sin x}{x}=1, \lim_{x \to 0 }\frac{tanx}{x}=1$

Also, 1 - cos 2 x= 2 sin²x, 1-cos x=2 sin²(x/2)