## Calculus, 10th Edition (Anton)

=$\frac{2}{5}$
Since the limit as $x$ approaches zero is $\frac{0}{0}$, L'Hopital's Rule can be used. Taking the derivative of the numerator and denominator (using chain rule) yields $$\lim\limits_{x \to 0}\frac{2cos(2x)}{5cos(2x)}=\frac{2cos(0)}{5cos(0)}=\frac{2}{5}$$