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I have tried changing limit to $\lim_{x\rightarrow0}$ and use some trigonometry identity ($\sin^2(x)+\cos^2(x) = 1$ and $\sin (x+\pi/2) = \cos(x)$) but doesn't work

I have no idea on how to do this now...

$$ \lim_{x\rightarrow{\frac\pi2 }} (\sec(x) \tan(x))^{\cos(x)} $$

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First observe that if $u=\cos x$:

$$ \lim_{x\to \pi/2} (\sec x \tan x)^{\cos x}=\lim_{x\to \pi/2} \left(\frac{1}{\cos^2x}\right)^ {\cos x}=\lim_{u\to 0} \left(\frac{1}{u^2}\right)^ {u} $$ Now just assume that $L=\lim_{u\to 0} \left(\frac{1}{u}\right)^ {2u}$. You get: $$ \ln L=\lim_{u\to 0} 2u\ln\left(\frac{1}{u}\right)=0 \implies L=1 $$


Remark: The last limit can be proved also without Hopital rule! To see this assume $y=\ln\left(\frac{1}{u}\right)$, then the limit turns out to be: $$ \lim_{y\to\infty}2ye^{-y} $$ To prove that the last limit is zero, we use the following inequality:

$$ 0<2ye^{-y}\leq \frac{2y}{1+y+y^2/2} $$ The limit of RHS goes to zero as $y\to\infty$.

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    $\begingroup$ He is taking $\lim_{x\to \pi/2} \sin x = 1$. I don't know how valid that might be though. $\endgroup$ – Parth Thakkar Sep 28 '13 at 16:58
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    $\begingroup$ I'm not sure if taking the partial limit $\,\sin x\xrightarrow[x\to\pi/2]{}1\;$ is valid in this case but, at any rate, there should be a proof of this. $\endgroup$ – DonAntonio Sep 28 '13 at 17:10
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    $\begingroup$ @DonAntonio, exactly my point. And experimentX, I agree that what you said holds true generally, however, I've come across certain limits where such a thing results only in absurdity. (Maybe I never understood the thing correctly!) For example: $$ \lim_{x\to 0} \dfrac{\sin x - x} {x^3} \\ = \lim_{x\to 0} \left(\dfrac{\sin x} {x^3} - \dfrac 1 {x^2} \right) \\ = \lim_{x\to 0} \left(\dfrac{\sin x} {x}\dfrac 1 {x^2} - \dfrac 1 {x^2} \right) \\ = \lim_{x\to 0} \left(1\cdot \dfrac 1 {x^2} - \dfrac 1 {x^2} \right) = 0$$ $\endgroup$ – Parth Thakkar Sep 28 '13 at 17:19
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    $\begingroup$ @ParthThakkar given that you can change $u = \pi/2 - x \to 0 $, this basically turns into (this type) $\sin(u)^{\sin(u)} \approx u^u , u\to 0$. it's easy to guess ... but to make it independent of L'hopital you have to show that $\lim_{x\to 0} x^x = 1$ without L'hopital. $\endgroup$ – Santosh Linkha Sep 28 '13 at 17:41
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    $\begingroup$ @IndyZa Hint: Taylor expansion of $e^x$. $\endgroup$ – Arash Sep 28 '13 at 17:58
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Hint: $$\lim_{t\to 0} \sin t \ln \frac{\cos t}{\sin^2t} = -2 \lim_{t\to 0} (\sin t \cdot \ln \sin t) = 0.$$

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  • $\begingroup$ Don't understand why ln just appear in the equation.. $\endgroup$ – IndyZa Sep 28 '13 at 17:23
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    $\begingroup$ He took $\ln$ of the limit. So, the $0$ he got is for $\ln(\text{limit})$ and not $\text{limit}$ as such. $\endgroup$ – Parth Thakkar Sep 28 '13 at 17:28
  • $\begingroup$ Is it just $ln(y)=\lim_{t\to 0} \sin t \ln \frac{\cos t}{\sin^2t} = -2 \lim_{t\to 0} (\sin t \cdot \ln \sin t) = 0.$ ? So $ln(y) = 0 and y = e^0 = 1?$ $\endgroup$ – IndyZa Sep 28 '13 at 17:38
  • $\begingroup$ Yeah. There was a correction in Arash's answer just now. $\endgroup$ – Parth Thakkar Sep 28 '13 at 17:41
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$\lim_{x \to \pi/2}\left[\sec\left(x\right)\tan\left(x\right)\right]^{\cos\left(x\right)} = \lim_{x \to 0}x^{-2x} = \lim_{x \to 0}{\rm e}^{-2x\ln\left(x\right)}$.

Since $\lim_{x \to 0}\left[-2x\ln\left(x\right)\right] = -2\lim_{x \to 0}{\ln\left(x\right) \over 1/x} = -2\lim_{x \to 0}{1/x \over -1/x^{2}} = 0$, $\displaystyle{\color{#ff0000}{\large\lim_{x \to \pi/2}\left[\sec\left(x\right)\tan\left(x\right)\right]^{\cos\left(x\right)} = 1}}$

$\newcommand{\abs}[1]{\left\vert #1\right\vert}$ ${\bf\mbox{Without L'H$\hat{\rm o}$pital}}$: $$ \abs{x\ln\left(x\right)} \leq \abs{x\left(x - 1\right)} \to 0\quad\mbox{when}\quad x \to 0 $$

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    $\begingroup$ Without Hopital my friend :) $\endgroup$ – Arash Sep 28 '13 at 17:55
  • $\begingroup$ @Arash 0 k. I have to read carefully. Thanks. $\endgroup$ – Felix Marin Sep 28 '13 at 17:56

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