Question

Помогите решить предел

Answer 1

$tg2x-2tgx=\frac{2tgx}{1-tg^2x}-2tgx=2tgx\cdot (\frac{1}{1-tg^2x}-1)=2tgx\cdot \frac{1-1+tg^2x}{1-tg^2x}=\\\\=2tgx\cdot \frac{tg^2x}{1-\frac{sin^2x}{cos^2x}}=2\cdot \frac{tg^3x\cdot cos^2x}{cos^2x-sin^2x}=2\cdot \frac{sin^3x\cdot cos^2x}{cos^3x\cdot cos2x}=2\cdot \frac{sin^3x}{cosx\cdot cos2x} \\\\\\\\sin2x-2sinx=2sinx\cdot cosx-2sinx=2sinx\cdot (cosx-1)=\\\\=-2sinx(1-cosx)=-2sinx\cdot 2sin^2\frac{x}{2}=-4sinx\cdot sin^2\frac{x}{2}$

$\lim\limits _{x \to 0}\frac{tg2x-2tgx}{sin2x-2sinx}=\lim\limits _{x \to 0}\frac{2\, sin^3x}{cosx\cdot cos2x\cdot (-4sinx\cdot sin^2\frac{x}{2})}=\\\\=\lim\limits _{x \to 0}\frac{sin^2x}{-2\cdot cosx\cdot cos2x\cdot sin^2\frac{x}{2}}=\lim\limits _{x \to 0}\frac{(2sin\frac{x}{2}\cdot cos\frac{x}{2})^2}{-2\cdot cosx\cdot cos2x\cdot sin^2\frac{x}{2}}=\\\\=\lim\limits _{x \to 0}\frac{4\cdot cos^2\frac{x}{2}}{-2\cdot cosx\cdot cos2x}=\frac{4\cdot 1}{-2\cdot 1\cdot 1}=-2\\\\\star \; \; cos0=1\; \; \star$