# Evaluate $\lim_{x\rightarrow{\frac\pi2 }} (\sec(x) \tan(x))^{\cos(x)}$ without L'Hôpital's rule

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...

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

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 $y\to\infty$.

• He is taking $\lim_{x\to \pi/2} \sin x = 1$. I don't know how valid that might be though. – Parth Thakkar Sep 28 '13 at 16:58
• 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. – DonAntonio Sep 28 '13 at 17:10
• @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: $$\lim_{x\to 0} \dfrac{\sin x - x} {x^3} \\ = \lim_{x\to 0} \left(\dfrac{\sin x} {x^3} - \dfrac 1 {x^2} \right) \\ = \lim_{x\to 0} \left(\dfrac{\sin x} {x}\dfrac 1 {x^2} - \dfrac 1 {x^2} \right) \\ = \lim_{x\to 0} \left(1\cdot \dfrac 1 {x^2} - \dfrac 1 {x^2} \right) = 0$$ – Parth Thakkar Sep 28 '13 at 17:19
• @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. – Santosh Linkha Sep 28 '13 at 17:41
• @IndyZa Hint: Taylor expansion of $e^x$. – Arash Sep 28 '13 at 17:58

Hint: $$\lim_{t\to 0} \sin t \ln \frac{\cos t}{\sin^2t} = -2 \lim_{t\to 0} (\sin t \cdot \ln \sin t) = 0.$$

• Don't understand why ln just appear in the equation.. – IndyZa Sep 28 '13 at 17:23
• He took $\ln$ of the limit. So, the $0$ he got is for $\ln(\text{limit})$ and not $\text{limit}$ as such. – Parth Thakkar Sep 28 '13 at 17:28
• 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 $ln(y) = 0 and y = e^0 = 1?$ – IndyZa Sep 28 '13 at 17:38
• Yeah. There was a correction in Arash's answer just now. – Parth Thakkar Sep 28 '13 at 17:41

$\lim_{x \to \pi/2}\left[\sec\left(x\right)\tan\left(x\right)\right]^{\cos\left(x\right)} = \lim_{x \to 0}x^{-2x} = \lim_{x \to 0}{\rm e}^{-2x\ln\left(x\right)}$.

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}$pital}}$: $$\abs{x\ln\left(x\right)} \leq \abs{x\left(x - 1\right)} \to 0\quad\mbox{when}\quad x \to 0$$

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