Question

Evaluate $\underset{x\to 0}{lim}\frac{1-\cos 4x}{x^2}$

Answer 1

$\underset{x\to 0}{lim}\frac{1-\cos 4x}{x^2}$
$=\underset{x\to 0}{lim}\frac{2{\sin }^{2}2x}{x^2}$
$=\underset{x\to 0}{lim}\frac{8{\sin }^{2}2x}{(2x{)}^2}$
$=8\cdot \underset{x\to 0}{lim}(\frac{\sin 2x}{2x}{)}^2=8\cdot 1$
=$8$