Question

Calculate the following limits.
$\underset{x\to 0}{lim}\frac{\cos 4x-cos6x}{{\sin }^{2}5x}.$

Answer 1

$\underset{x\to 0}{lim}\frac{\cos 4x-\cos 6x}{{\sin }^{2}5x}$

$=\underset{x\to 0}{lim}\frac{2\sin 5x\sin x}{{\sin }^{2}5x}$

$=\underset{x\to 0}{lim}\frac{2\sin x}{\sin 5x}$

$=\underset{x\to 0}{lim}\frac{2\sin x}{x}\frac{5x}{\sin 5x}\frac{1}{5}$

$=\frac{2}{5}$