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I'm a bit rusty with limits and derivatives at the moment. I was doing L'hosp on another problem when I got stuck here. $$\lim_{x \to 0} \dfrac{\dfrac{\mathrm d}{\mathrm d x} (e^{\large \sec x})}{\dfrac{\mathrm d}{\mathrm d x} (e^{\large\sec 2x})}$$ Further L'hosp is a mess. Care to continue, my jolly fellows?

amWhy
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Nick
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    Is the motivation for your limit the calculation of this limit?

    $$\lim_{x\to 0} \frac{e^{\sec x}}{e^{\sec 2x}}$$

    Because if so, this limit can (and should) be calculated without l'Hopital's rule.

     – Simon S Dec 18 '14 at 14:20
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    ...indeed, MUST be calculated without l'Hopital's rule. – vadim123 Dec 18 '14 at 14:22
  • @SimonS And even without a $\lim$ because $\sec 0 = 1$, so the whole thing is $1$. – GDumphart Dec 18 '14 at 14:23
  • @SimonS: It was initially something more complicated than that but yes, I did reduce it down to $$\lim_{x\to 0} \frac{1-e^{\sec x}}{1 - e^{\sec 2x}}$$ – Nick Dec 18 '14 at 14:25
  • @Nick In that case, the $\lim$ makes sense again. – GDumphart Dec 18 '14 at 14:26
  • Yes, what @GDumphart said. So calculate it out. The derivative of $\sec x$ is $\sec x \tan x$. – Simon S Dec 18 '14 at 14:26
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    @Nick Please neither use displaystyle maths (\displaystyle, \dfrac) nor MathJax-only titles. Both are discouraged. The former because it costs much more front-page space and the latter due to technical reasons (Users unable to invoke the browser context-menu on the link) – AlexR Dec 18 '14 at 14:28
  • sec(0) = 1. So the limit you want can be calculated by directly plugging 0 for x and the answer would be 1 – chandu1729 Dec 18 '14 at 14:29
  • @chandu1729: Actually, my initial limit had $e^{\sin x} -e^{\sin2x}$. So, I thought, since it evaluated to 1, all I have left is the the other guy which you now say is 1... So, I'm pretty sure I messed up by splitting the limits and evaluating. – Nick Dec 18 '14 at 14:31
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    Please post your initial problem :) – chandu1729 Dec 18 '14 at 14:32
  • @chandu1729: Actually, I've found a way to solve that problem. The limit in my post is much more interesting to me now. I've never actually encountered a problem with derivatives explicitly stuck in it. This was, in a way, worth the ask. – Nick Dec 18 '14 at 14:44
  • @SimonS: Ah, silly me. But I do have to admit, this was a nice question that I accidently cooked up. I'm glad for asking it :D – Nick Dec 18 '14 at 15:07

2 Answers2

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Let $f(x) = e^{\sec x}$. Then you are looking for $$\displaystyle\lim_{x \to 0} \dfrac{\dfrac{\mathrm d}{\mathrm d x} (e^{\sec x})}{\dfrac{\mathrm d}{\mathrm d x} (e^{\sec 2x})} = \lim_{x \to 0}{f'(x) \over 2 f'(2x)}$$ Note that $f'(x) = \tan x \sec x \,e^{\sec x}$, so that $f'(0) = 0$. This means the limit is of $0/0$ form, so we use L'hopital, and the limit becomes $$= \lim_{x \to 0}{f''(x) \over 4 f''(2x)}$$ Here $f''(x) = \sec^3 x \, e^{\sec x} + \tan^2 x \sec x \, e^{\sec x} + (\tan x \sec x)^2\, e^{\sec x}$, and $f''(0) = 1 \neq 0$. Hence the limit is ${1 \over 4}$.

Zarrax
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Executing the derivatives we get: $$ \frac{1}{2}\left(\lim_{x\rightarrow 0}\left( \frac{\frac{\sin(x)}{\cos^2(x)}e^{1/\cos(x)}}{\frac{\sin(2x)}{\cos^2(2x)}e^{1/\cos(2x)}} \right)\right) $$ $=$ $$ \frac{1}{2}\lim_{x\rightarrow 0}\left(\frac{\sin(x)}{\sin(2x)}\right)\left(\lim_{x\rightarrow 0}\left( \frac{\frac{1}{\cos^2(x)}e^{1/\cos(x)}}{\frac{1}{\cos^2(2x)}e^{1/\cos(2x)}} \right)\right) $$ In the last part we have factored out a surely converging part.

Using that $e^x\approx 1+x$ , $\sin(x)\approx x$ , $\cos(x)\approx 1$ , $e^{1/\cos(ax)}\approx 1+\frac{1}{\cos(ax)}\approx2$ at $x$=0 we can rewrite this as

$$ \frac{1}{2}\left(\lim_{x\rightarrow 0}\left( \frac{\frac{x}{1}2}{\frac{2 x}{1}2} \right)\right)=\frac{1}{4} $$

tired
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