Question

Evaluate $\underset{x\to 0}{lim}\frac{\left(e^{2x}-1\right)\left(1-\cos 4x\right)}{x^3}$

Answer 1

${lim}_{x\to 0}\frac{(e^{2x}-1)(1-\cos 4x)}{x^3}={lim}_{x\to 0}\frac{e^{2x}-1}{x}.{lim}_{x\to 0}\frac{1-\cos 4x}{x^2}$

$={lim}_{x\to 0}\frac{2e^{2x}}{1}.{lim}_{x\to 0}\frac{e^{2x}-1}{x}.{lim}_{x\to 0}\frac{4\sin 4x}{2x}$

$=2.{lim}_{x\to 0}\frac{16\cos 4x}{2}$

$=2\times 8=16$