# Math Help - limit as h approaches 0

1. ## limit as h approaches 0

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

2. Originally Posted by DINOCALC09
What is the

limit of [ tan(pi/6+h) - tan(pi/6) ] / h

as h approaches 0
First, $\tan (x+y) = \frac{\tan x + \tan y}{1-\tan x \tan y}$.

This means, $\tan \left( \frac{\pi}{6} + h \right) = \frac{\tan \frac{\pi}{6} + \tan h}{1 - \tan \frac{\pi}{6} \cdot \tan h} = \frac{\frac1{\sqrt{3}} + \tan h}{1 - \tan h \cdot \frac1{\sqrt{3}}}$

Thus, $\tan \left( \frac{\pi}{6} + h \right) - \tan \frac{\pi}{6} = \frac{\frac1{\sqrt{3}} + \tan h}{1 - \tan h \cdot \frac1{\sqrt{3}}} - \frac1{\sqrt{3}} = \frac{\frac 43\tan h}{1 - \tan h \cdot \sqrt{3}}$

If you divide this by $h$, you get, $\frac{\frac43\frac{\tan h}{h}}{1 - \tan h \cdot \sqrt{3}}$.

Take $h\to 0$ and use the fact that $\lim_{h\to 0}\frac{\tan h}{h} = 1$ and we get that the limit is $\frac{4}{3}$.

3. One might note that:
$\lim _{h \to 0} \left[ {\frac{{\tan \left( {\frac{\pi }{6} + h} \right) - \tan \left( {\frac{\pi }{6}} \right)}}{h}} \right] = \sec ^2 \left( {\frac{\pi }{6}} \right)$

4. Originally Posted by Plato
One might note that:
$\lim _{h \to 0} \left[ {\frac{{\tan \left( {\frac{\pi }{6} + h} \right) - \tan \left( {\frac{\pi }{6}} \right)}}{h}} \right] = \sec ^2 \left( {\frac{\pi }{6}} \right)$
It did not seem to me that he knew what a derivative is.

5. Originally Posted by ThePerfectHacker
It did not seem to me that he knew what a derivative is.
I bet you that is what the question is about.
It is just too perfectly setup.