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Math Help - [SOLVED] limit w/tan

  1. #1
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    [SOLVED] limit w/tan

    limit as x approaches pi/2 of (tan x)^(x-pi/2) I don't even know where to start.
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  2. #2
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    Quote Originally Posted by Amberosia32 View Post
    limit as x approaches pi/2 of (tan x)^(x-pi/2) I don't even know where to start.

    \left(\tan x\right)^{x-\frac{\pi}{2}}=e^{\left(x-\frac{\pi}{2}\right)\ln\tan x} , so using continuity of the exponential formula, apply L'Hospital's rule to:

    \lim_{x\to \pi\slash 2}\left(x-\frac{\pi}{2}\right)\ln\tan x=\lim_{x\to \pi\slash 2}\frac{\ln\tan x}{\frac{1}{x-\pi\slash 2}} .

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    \left(\tan x\right)^{x-\frac{\pi}{2}}=e^{\left(x-\frac{\pi}{2}\right)\ln\tan x} , so using continuity of the exponential formula, apply L'Hospital's rule to:

    \lim_{x\to \pi\slash 2}\left(x-\frac{\pi}{2}\right)\ln\tan x=\lim_{x\to \pi\slash 2}\frac{\ln\tan x}{\frac{1}{x-\pi\slash 2}} .

    Tonio
    Where did the e come from? I really don't understand this at all, sorry. I'm sure it's something I should know. Or did you just use it to use ln to bring down the power?
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  4. #4
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    well you have tan(pi/2)^0, correct? Well tan is sin/cos, and cos of pi/2 is 0, which would make it undefined. But it's APPROACHING pi/2, not at it. look at it graphically, pi/2 is an asymptote of tanx and to the left of the asymptote f(x) approaches infinity.

    So you're really looking at [sin(pi/2)/cos(pi/2)]^0 or [1/inf]^0 or 0^0. This is an indeterminite form of a limit and I don't know if you've learned L'Hopitals rule yet but it needs to be applied here.

    We write y=tanx^(x-pi/2) and take the natural log of both sides. lny=(x-pi/2)lntanx we rewrite this to get it to the form 0/0 so we can l'hop it.

    lny= lntanx/(x-pi/2)^-1 which is 0/0 as x>pi/2.

    now we take the derivative of top and bottom to lhop it.
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  5. #5
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    what he did with the e is just to exponentiate at the end. becuase you'll be left with lny= whatever so you change it to e^lny=e^whatever which is y=e^whatever

    I think the answers 1 btw cuz whatever should be 0 after the lhop and e^0 is 1.
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  6. #6
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    Quote Originally Posted by Jph93 View Post
    what he did with the e is just to exponentiate at the end. becuase you'll be left with lny= whatever so you change it to e^lny=e^whatever which is y=e^whatever

    I think the answers 1 btw cuz whatever should be 0 after the lhop and e^0 is 1.
    Thanks for explaining the e thing. lol. I feel kinda slow now. You're a lifesaver.
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  7. #7
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    Quote Originally Posted by Amberosia32 View Post
    Where did the e come from? I really don't understand this at all, sorry. I'm sure it's something I should know. Or did you just use it to use ln to bring down the power?

    Well, yes: i think you should know it, but it never minds: you'll now it if you continue reading.

    For any a\,,\,b\in\mathbb{R}\,,\,a>0\,,\,\,a^b=e^{b\ln a} . This follows at once from the very definition of logarithm which I urge to read (or hopefully: re-read) and re-understand.

    Tonio
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