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Math Help - lim_(x->0^+) (tan(3 x))^x

  1. #1
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    lim_(x->0^+) (tan(3 x))^x

    how would i go about doing this problem?

    lim_(x->0^+) (tan(3 x))^x

    I am thinking of using lim_(x->0^+) y =?

    lny = xln(tan(3x))

    lntan(3x)/ (x^-1)

    3(sec(3x)^2 / (-x^-2)


    am i on the right path?
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  2. #2
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    Re: lim_(x->0^+) (tan(3 x))^x

    derivative of \ln[\tan(3x)] is wrong
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: lim_(x->0^+) (tan(3 x))^x

    What you're doing is good, write the limit as:
    \lim_{x\to 0^{+}} [\tan(x)]^{x}=\lim_{x\to 0^{+}} e^{x\cdot \ln\left[\tan(x)]}

    Now you can just determine the limit, you don't have to use l'Hopitals rule.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: lim_(x->0^+) (tan(3 x))^x

    Out of curiosity: using \tan \epsilon \sim \epsilon\;(\epsilon\to 0^+) we have \lim_{x\to 0^+}\dfrac{\tan 3x}{3x}=1 so

    L=\displaystyle\lim_{x \to 0^+}\left(\tan 3x}\right)^x= \displaystyle\lim_{x \to 0^+}\left(\dfrac{\tan 3x}{3x}\cdot 3x\right)^x=

    \displaystyle\lim_{x \to 0^+}\left(\dfrac{\tan 3x}{3x}\right)^x\cdot \displaystyle\lim_{x \to 0^+}\left( 3x\right)^x= \displaystyle\lim_{x \to 0^+}\left( 3x\right)^x =\ldots
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: lim_(x->0^+) (tan(3 x))^x

    Quote Originally Posted by NeoSonata View Post
    how would i go about doing this problem?

    lim_(x->0^+) (tan(3 x))^x

    I am thinking of using lim_(x->0^+) y =?

    lny = xln(tan(3x))

    lntan(3x)/ (x^-1)

    3(sec(3x)^2 / (-x^-2)


    am i on the right path?
    Yes, You are!... if You consider a more general complex variable z You have...

    \tan 3 z = 3\ z\ \prod_{n=1}^{\infty} \frac{1-\frac{9\ z^{2}}{n^{2} \pi^{2}}}{1-\frac{9\ z^{2}}{(2 n-1)^{2} \pi^{2}}} (1)

    ... and from (1) You derive...

    \ln \tan 3z = \ln 3 + \ln z + \sum_{n=1}^{\infty} \ln \{1-\frac{9\ z^{2}}{n^{2} \pi^{2}}\} - \sum_{n=1}^{\infty} \ln \{1-\frac{9\ z^{2}}{(2 n-1)^{2} \pi^{2}}\} (2)

    Now from (2) is easy to derive that is...

    \lim_{z \rightarrow 0} z\ \ln \tan 3 z = 0 (3)

    ... so that...

    \lim_{z \rightarrow 0} (\tan 3 z)^{z} = 1 (4)

    It is remarkable the fact that the limit (3) and (4) are independent from how z tends to 0...

    Kind regards

    \chi \sigma
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