Question

${lim}_{x\to 0}\frac{{(\sin x-\tan x)}^2+{(1-\cos 2x)}^4+x^5}{7{\tan }^{7}x+{\sin }^{6}x+2{\sin }^{5}x}$ is equals to

Answer 1

${lim}_{x\to 0}\frac{{(\sin x-\tan x)}^2+{(1-\cos 2x)}^4+x^5}{7{\tan }^{7}x+{\sin }^{6}x+2{\sin }^{5}x}$

$={lim}_{x\to 0}\frac{{\sin }^{2}x+{\tan }^{2}x-2\sin x\tan x+{\left(2{\sin }^{2}x\right)}^4+x^5}{7{\tan }^{7}x+{\sin }^{6}x+2{\sin }^{5}x}$

$={lim}_{x\to 0}\frac{\frac{{\sin }^{2}x}{x^2}\times x^2+\frac{{\tan }^{2}x}{x^2}\times x^2-2\frac{\sin x}{x}\times \frac{\tan x}{x}\times x^2+\frac{16{\sin }^{8}x}{x^8}\times x^8+x^5}{7\frac{{\tan }^{7}x}{x^7}\times x^7+\frac{{\sin }^{6}x}{x^6}\times x^6+2\frac{{\sin }^{5}x}{x^5}\times x^5}$

$={lim}_{x\to 0}\frac{1\times x^2+1\times x^2-2\times 1\times 1\times x^2+16\times 1\times x^8+x^5}{7\times 1\times x^7+x^6+2x^5}$

$={lim}_{x\to 0}\frac{16x^{13}}{7x^7+x^6+2x^5}$

$={lim}_{x\to 0}\frac{16x^{13}}{x^5(7x^2+x+2)}$

$={lim}_{x\to 0}\frac{16x^{13-5}}{(7x^2+x+2)}$

$={lim}_{x\to 0}\frac{16x^8}{(7x^2+x+2)}$

$=\frac{0}{0+0+2}=0$