Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.5 L'Hopital's Rule; Indeterminate Forms - Exercises Set 6.5 - Page 448: 8

=$\frac{2}{5}$

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 $$\lim\limits_{x \to 0}\frac{2cos(2x)}{5cos(2x)}=\frac{2cos(0)}{5cos(0)}=\frac{2}{5}$$