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Limit with Tan()

  1. Jul 21, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]\displaystyle \lim_{x\rightarrow \frac{\pi}{4}}\left(\tan x\right)^{\tan 2x}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    let [tex]\frac{\pi}{4}-x = t\Leftrightarrow x=\frac{\pi}{4}-t[/tex] and [tex]t\rightarrow 0[/tex] [tex]\lim_{t\rightarrow 0}\left(\tan \left(\frac{\pi}{4}-t\right)\right)^{\tan \left(\frac{\pi}{2}-2t\right)}[/tex]
    [tex]\lim_{t\rightarrow 0}\left(\frac{1-\tan t}{1+\tan t}\right)^{\frac{1}{\tan 2t}}[/tex]
    as [tex]t\rightarrow 0,\tan t\approx t,\tan 2t\approx 2t[/tex]
    [tex]\lim_{t\rightarrow 0}\left(\frac{1- t}{1+ t}\right)^{\frac{1}{2t}}[/tex]
    after that how can i solve
  2. jcsd
  3. Jul 21, 2011 #2


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    hi juantheron! :smile:

    lim (1/(1 - t)) = 1 + … ? :wink:
  4. Jul 21, 2011 #3


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    Re: limit

    Use the fact that [itex]\displaystyle \lim f(x)=e^{\lim\ln(f(x))}[/itex].

    In evaluating lim(ln(f(x)), use L'Hôpital's rule.
  5. Jul 22, 2011 #4
    Re: limit

    SammyS's method is much closer to the most effective method I can think of.

    AB = (e ln A)B = e B * ln A

    Note: tan x is always positive in the vicinity we are interested in, so ln A is well defined.

    Apply L'Hospital's rule to the power.

    For oo * 0 forms, we need [itex]\frac {oo} {\frac{1}{0}}[/itex] or [itex]\frac{0} {\frac{1}{oo}}[/itex] before we can apply L'Hospital's rule.
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