finding limit

**Sambit** · January 23rd 2011, 12:09 AM

What is the value of $\lim_{x \to \infty} (\frac{\pi}{2}-\tan^{-1} x)^{1/x}?$

**Prove It** · January 23rd 2011, 12:22 AM

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

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

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

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

This limit goes to $\displaystyle \frac{-\infty}{\infty}$ so you can use L'Hospital's Rule...

**Soroban** · January 23rd 2011, 04:55 AM

Hello, Sambit!

Another approach . . .

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

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

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

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

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

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

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

**Prove It** · January 23rd 2011, 05:05 AM

Originally Posted by Soroban

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

**Krizalid** · January 23rd 2011, 10:34 AM

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 $e^0=1.$

**Sambit** · January 23rd 2011, 10:17 PM

Thanks to all...

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