Question

Lim x tends to o xtanx/1-cosx

Answer 1

$\lim_{n \to 0} \frac{xtanx}{1-cosx}$

We know, 1 - cos2A = 2sin²A , use this concept here,
So, $\lim_{n \to 0} \frac{xtanx}{2sin^2\frac{x}{2}}$

now, $\lim_{n \to 0} \frac{xtanx}{2sin^2\frac{x}{2}}\\\\= \lim_{n \to 0}\frac{tanx}{x}.\frac{x^2}{\frac{2sin^2x/2}{x^2/4}.x^2/4}\\\\ =\lim_{n \to 0}1.\frac{4x^2}{2x^2}\\\\=2$