What is the
limit of [ tan(pi/6+h) - tan(pi/6) ] / h
as h approaches 0
A) sqrt(3) / 3
B) 4/3
C) sqrt(3)
D) 0
E. 3/4
First, $\displaystyle \tan (x+y) = \frac{\tan x + \tan y}{1-\tan x \tan y}$.
This means, $\displaystyle \tan \left( \frac{\pi}{6} + h \right) = \frac{\tan \frac{\pi}{6} + \tan h}{1 - \tan \frac{\pi}{6} \cdot \tan h} = \frac{\frac1{\sqrt{3}} + \tan h}{1 - \tan h \cdot \frac1{\sqrt{3}}}$
Thus, $\displaystyle \tan \left( \frac{\pi}{6} + h \right) - \tan \frac{\pi}{6} = \frac{\frac1{\sqrt{3}} + \tan h}{1 - \tan h \cdot \frac1{\sqrt{3}}} - \frac1{\sqrt{3}} = \frac{\frac 43\tan h}{1 - \tan h \cdot \sqrt{3}}$
If you divide this by $\displaystyle h$, you get, $\displaystyle \frac{\frac43\frac{\tan h}{h}}{1 - \tan h \cdot \sqrt{3}}$.
Take $\displaystyle h\to 0$ and use the fact that $\displaystyle \lim_{h\to 0}\frac{\tan h}{h} = 1$ and we get that the limit is $\displaystyle \frac{4}{3}$.