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Math Help - finding limit

  1. #1
    Senior Member Sambit's Avatar
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    Question finding limit

    What is the value of \lim_{x \to \infty} (\frac{\pi}{2}-\tan^{-1} x)^{1/x}?
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  2. #2
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    \displaystyle \lim_{x \to \infty}\left(\frac{\pi}{2} - \arctan{x}\right)^{\frac{1}{x}} = \lim_{x \to \infty}e^{\ln{\left[\left(\frac{\pi}{2} - \arctan{x}\right)^{\frac{1}{x}}\right]}}

    \displaystyle = e^{\lim_{x \to \infty}\ln{\left[\left(\frac{\pi}{2} - \arctan{x}\right)^{\frac{1}{x}}\right]}}

    \displaystyle = e^{\lim_{x \to \infty}\frac{1}{x}\ln{\left(\frac{\pi}{2} - \arctan{x}\right)}}

    \displaystyle = e^{\lim_{x\to \infty}\frac{\ln{\left(\frac{\pi}{2} - \arctan{x}\right)}}{x}}

    This limit goes to \displaystyle \frac{-\infty}{\infty} so you can use L'Hospital's Rule...
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  3. #3
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    Hello, Sambit!

    Another approach . . .


    \displaystyle\text{Find the value of: }\;\lim_{x \to \infty} \left(\tfrac{\pi}{2}-\tan^{-1}\!x\right)^{\frac{1}{x}}

    \text{Let: }\,y \:=\:\left(\frac{\pi}{2} - \tan^{-1}\!x\right)^{\frac{1}{x}}

    \text{Take logs: }\;\ln y \;=\;\ln\left(\frac{\pi}{2} - \tan^{-1}\!x\right)^{\frac{1}{x}} \;=\;\dfrac{\ln(\frac{\pi}{2} - \tan^{-1}\!x)}{x}


    \displaystyle\text{Then: }\;\lim_{x\to\infty} \ln y \;=\;\lim_{x\to\infty}\frac{\ln(\frac{\pi}{2} - \tan^{-1}\!x)}{x} \;\to\;-\frac{\infty}{\infty}


    \text{Apply l'Hopital: }\;\dfrac{\frac{\pi}{2} - \tan^{-1}\!x}{x} \;\Rightarrow\; \dfrac{\text{-}\frac{1}{1+x^2}}{1} \;=\;\dfrac{\text{-}1}{1+x^2}


    \displaystyle \text{Hence: }\;\lim_{x\to\infty}\ln y \;=\;\lim_{x\to\infty}\frac{\text{-}1}{1+x^2} \;=\;0


    \displaystyle\text{Therefore: }\;\lim_{x\to\infty}y \;=\;e^0 \;=\;1

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, Sambit!

    Another approach . . .



    \text{Let: }\,y \:=\:\left(\frac{\pi}{2} - \tan^{-1}\!x\right)^{\frac{1}{x}}

    \text{Take logs: }\;\ln y \;=\;\ln\left(\frac{\pi}{2} - \tan^{-1}\!x\right)^{\frac{1}{x}} \;=\;\dfrac{\ln(\frac{\pi}{2} - \tan^{-1}\!x)}{x}


    \displaystyle\text{Then: }\;\lim_{x\to\infty} \ln y \;=\;\lim_{x\to\infty}\frac{\ln(\frac{\pi}{2} - \tan^{-1}\!x)}{x} \;\to\;-\frac{\infty}{\infty}


    \text{Apply l'Hopital: }\;\dfrac{\frac{\pi}{2} - \tan^{-1}\!x}{x} \;\Rightarrow\; \dfrac{\text{-}\frac{1}{1+x^2}}{1} \;=\;\dfrac{\text{-}1}{1+x^2}


    \displaystyle \text{Hence: }\;\lim_{x\to\infty}\ln y \;=\;\lim_{x\to\infty}\frac{\text{-}1}{1+x^2} \;=\;0


    \displaystyle\text{Therefore: }\;\lim_{x\to\infty}y \;=\;e^0 \;=\;1

    Actually it's the exact same approach
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  5. #5
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    the limit is \displaystyle\underset{x\to 0}{\mathop{\lim }}\,{{x}^{\tan x}} after substitution t=\dfrac{\pi }{2}-\arctan x, now for x\ne0 we have \displaystyle{{x}^{\tan x}}=\exp \left( \frac{\sin x}{\cos x}\cdot \ln x \right)=\exp \left( \frac{\sin x}{x}\cdot x\ln (x)\cdot \frac{1}{\cos x} \right), as x\to0^+ the limit is e^0=1.
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  6. #6
    Senior Member Sambit's Avatar
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    Thanks to all...
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