Answer
$c=\frac{4}{3}$
Work Step by Step
$\lim\limits_{x \to 0}\frac{4x-2sin(2x)}{2x^{3}}$=$\frac{0}{0}$
$\lim\limits_{x \to 0}\frac{4-4cos(2x)}{6x^{2}}$=$\frac{0}{0}$
$\lim\limits_{x \to 0}\frac{8sin(2x)}{12x}$=$\frac{0}{0}$
$\lim\limits_{x \to 0}\frac{16cos(2x)}{12}$=$\frac{16cos(2(0))}{12}$=$\frac{16}{12}$=$\frac{4}{3}$
