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

$\begin{array}{l}=\underset{x\to 0}{lim} \frac{(\sin x-\tan x{)}^2+(1-\cos 2x{)}^4+x^5}{7{\tan }^{7}x+{\sin }^{6}x+2{\sin }^{5}x}\\ =\underset{x\to 0}{lim} \frac{{\sin }^{2}x{\left(\frac{1-{\cos }^{2}x}{{\cos }^{2}x}\right)}^2+16{\sin }^{8}x+x^5}{7\frac{{\sin }^{7}x}{{\cos }^{7}x}+{\sin }^{6}x+2{\sin }^{5}x}\\ \underset{x\to 0}{=lim} \frac{{\left(\frac{\sin x}{x}\right)}^5\sin x\cdot {\sec }^{4}x+16{\left(\frac{\sin x}{x}\right)}^5{\sin }^{3}x+1}{{\left.7\frac{\sin x}{x}\right)}^5{\sin }^{2}x{\sec }^{7}x+{\left(\frac{\sin x}{x}\right)}^5\sin x+2{\left(\frac{\sin x}{x}\right)}^5}\\ =\frac{1\times 0\times 1+16\times 1\times 0+1}{7\times 1\times 0\times 1+1\times 0+2\times 1}=\frac{1}{2}\cdot \end{array}$