Question

Prove that
${lim}_{x\to 0}\frac{2\sin x-\sin 2x}{x^2}$

Answer 1

$\underset{x\to 0}{lim}\frac{2\sin x-{\sin }^{2}x}{x^2}$

$\underset{x\to 0}{lim}\frac{2\sin x(1-\cos x)}{x^2}$

$\underset{x\to 0}{lim}\frac{2\sin x\cdot 2{\sin }^{2}x/2}{4\cdot (x/2{)}^2}$

$\underset{x\to 0}{lim}\sin \underset{x\to 0}{lim}\frac{{\sin }^{2}x/2}{(x/2{)}^2}$

$=0\times 1$

$=0$.