Question

Evaluate:${lim}_{x\to 0}\frac{{\sin }^{2}2x}{{\sin }^{2}4x}$

Answer 1

$={lim}_{x\to 0}\frac{\frac{{\sin }^{2}2x}{{\left(2x\right)}^2}\times 4x^2}{\frac{{\sin }^{2}4x}{{\left(4x\right)}^2}\times 16x^2}$

$={lim}_{x\to 0}\frac{4x^2}{16x^2}=\frac{1}{4}$