Question

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

Answer 1

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

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

$=\frac{1\times 4x^2}{1\times 16x^2}$ since $\frac{\sin \theta }{\theta }=1$

$=\frac{4}{16}=\frac{1}{4}$ on simplification.