## anonymous 5 years ago lim=2tan(x/3)/x as x goes to 0

1. anonymous

You must use L'Hopital's rule. Do you know how?

2. anonymous

NO

3. anonymous

$\lim_{x \rightarrow 0} \frac{2\tan(\frac{x}{3})}{x}$ you can use the squeeze theorem here and get : $\frac{-\pi}{2} < \tan(\frac{x}{3}) < \frac{\pi}{2}$ $\frac{\frac{-2\pi}{2}}{x} < \frac{2\tan(\frac{x}{3})}{x} <\frac{ \frac{2\pi}{2}}{x}$ $\lim_{x \rightarrow 0} \frac{-\pi}{x} = \lim_{x \rightarrow 0}\frac{\pi}{x} = \lim_{x \rightarrow 0} \frac{2\tan(\frac{x}{3})}{x} = \infty$ ^_^ correct me if I'm wrong

4. anonymous

$\lim_{x\rightarrow 0}\frac{2\tan\frac{x}{3}}{x}= \lim_{x\rightarrow 0}\frac{\frac{2}{3}\cdot\tan\frac{x}{3}}{\frac{x}{3}}=\frac{2}{3}\cdot\lim_{y\rightarrow 0}\frac{\tan y}{y}= \frac{2}{3}\cdot\lim_{y\rightarrow 0}\frac{1}{\cos^2y}=\frac{2}{3}$

5. anonymous

hmm, if you used l'hopital's rule and got that, then there must've been a mistake in my calculations ? ^_^

6. anonymous

;-)

7. anonymous

hmm, but I'm sure of my steps though

8. anonymous

You derive the numerator and the denominator independently until you end up with an answer other than 0/0 or infinity/infinity. The limit is 2/3.

9. anonymous

but yomary is not aware of this way, so I tried the squeze theorem, but somehow it gives me a different answer ,hmm

10. anonymous

$-\frac{\pi}{2}<\tan\frac{x}{3}<\frac{\pi}{2}$ ???

11. anonymous

well, tanx is , can't tan(x/3) be too?