Question

Find ${lim}_{x\to 0}\frac{\sin x-\tan x}{x^3}$

Answer 1

${lim}_{x\to 0}\frac{sinx-tanx}{x^3}$

$={lim}_{x\to 0}\frac{sinx(1-\frac{1}{cosx})}{x^3}$

$={lim}_{x\to 0}\frac{sinx}{x}\left(\frac{-(1-cosx)}{x^{2}cosx}\right)$

$={lim}_{x\to 0}\frac{sinx}{x}.{lim}_{x\to 0}\left(\frac{-2si{n}^2\frac{x}{2}}{\frac{x^2}{4}\times 4cosx}\right)$

$=1\times \frac{-2}{4}=\frac{-1}{2}$