Question

${lim}_{x\to 0}\frac{1-cos4x}{x^2}$

Answer 1

${lim}_{x\to 0}\frac{1-cos4x}{x^2}$

$={lim}_{x\to 0}\frac{2si{n}^{2}2x}{x^2}=2{lim}_{x\to 0}{\left(\frac{sin2x}{x}\right)}^2$

$=2{lim}_{x\to 0}{\left(\frac{sin2x}{2x}\right)}^2\times (2{)}^2$

$2\times 1\times 4=8$