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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    You must use L'Hopital's rule. Do you know how?

  2. anonymous
    • 5 years ago
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    NO

  3. anonymous
    • 5 years ago
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    \[\lim_{x \rightarrow 0} \frac{2\tan(\frac{x}{3})}{x}\] 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. nikvist
    • 5 years ago
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    \[\lim_{x\rightarrow 0}\frac{2\tan\frac{x}{3}}{x}= \lim_{x\rightarrow 0}\frac{\frac{2}{3}\cdot\tan\frac{x}{3}}{\frac{x}{3}}=\frac{2}{3}\cdot\lim_{y\rightarrow 0}\frac{\tan y}{y}= \frac{2}{3}\cdot\lim_{y\rightarrow 0}\frac{1}{\cos^2y}=\frac{2}{3}\]

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

  6. anonymous
    • 5 years ago
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    ;-)

  7. anonymous
    • 5 years ago
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    hmm, but I'm sure of my steps though

  8. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    but yomary is not aware of this way, so I tried the squeze theorem, but somehow it gives me a different answer ,hmm

  10. nikvist
    • 5 years ago
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    \[-\frac{\pi}{2}<\tan\frac{x}{3}<\frac{\pi}{2}\] ???

  11. anonymous
    • 5 years ago
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    well, tanx is , can't tan(x/3) be too?

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