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Thread: finding limit

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

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

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

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

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

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

    Another approach . . .


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

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

    $\displaystyle \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} $


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


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


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


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

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

    Another approach . . .



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

    $\displaystyle \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} $


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


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


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


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

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