Question

lim tanx - sinx / x ( 1 - cos2x )
x - > 0

Answer 1

Solution:

$\lim_{x \to 0} \frac{tanx-sinx}{x(1-cos2x)}\\\\  \lim_{x \to 0} \frac{tanx(1-cosx)}{x(1-cos2x)}\\\\  \lim_{x \to 0}\frac{tanx}{x} \lim_{x \to 0}\frac{1-cosx}{1-cos2x}\\\\=1 \times \lim_{x \to 0} \frac{2sin^2\frac{x}{2}}{2sin^2x}\\\\\lim_{x \to 0} \frac{\frac{sin^2\frac{x}{2}}{4\times(\frac{x}{2})^2}}{\frac{{sin^2x}}{x^2}} \lim_{x \to 0}\frac{tanx}{x}=1 and   \lim_{x \to 0}\frac{sinx}{x}=1 \\\\ lim_{x \to 0} \frac{1}{4}\times 1=\frac{1}{4}$

Keep in mind

→1 -cos 2x=2sin²x

→1 -cos x$=2sin^2{\frac{x}{2}$

Answer 2

in this copy answer we see it