Chapter 4 - Review - Exercises - Page 359: 8

$\lim\limits_{x \to 0}\frac{tan~4x}{x+sin~2x}= \frac{4}{3}$

$\lim\limits_{x \to 0}\frac{tan~4x}{x+sin~2x} = \frac{0}{0}$ We can use L'Hospital's Rule: $\lim\limits_{x \to 0}\frac{tan~4x}{x+sin~2x}$ $=\lim\limits_{x \to 0}\frac{\frac{4}{cos^2~4x}}{1+2~cos~2x}$ $=\frac{4}{1+2}$ $= \frac{4}{3}$