Question

lim=2tan(x/3)/x as x goes to 0

Answer 1

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

Answer 2

NO

Answer 3

\[\lim_{x \rightarrow 0} \frac{2\tan(\frac{x}{3})}{x}\]

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

Answer 4

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

Answer 5

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

Answer 6

;-)

Answer 7

hmm, but I'm sure of my steps though

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

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