If you try to find the $\lim_{x\to \infty}\frac{\sqrt{x^2+1}}{x}$ using L’Hôpital’s rule, you’ll find that it flip-flops back and forth between $\frac{\sqrt{x^2+1}}{x}$ and $\frac{x}{\sqrt{x^2+1}}$.

Are there other expressions that do a similar thing when L’Hôpital’s rule is applied to them? I already know that this applies to any fraction of the form $\frac{\sqrt{x^{2n}+c}}{x^n}$.

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    $\begingroup$ Your title made it sound like you had found a use for this behavior ... $\endgroup$ – abiessu Jun 26 '15 at 2:54
  • $\begingroup$ sorry I'll change it $\endgroup$ – Zach L Jun 26 '15 at 2:54
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    $\begingroup$ For this particular one you can square it first, use L'Hospital, and then take the square root, which gives $1$, as expected. Interesting question, though. $\endgroup$ – Ivo Terek Jun 26 '15 at 2:55
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    $\begingroup$ If I have understood, you are trying to solve the differential equation $\frac{f'(x)}{g'(x)}=\frac{g(x)}{f(x)}$, yes? $\endgroup$ – ajotatxe Jun 26 '15 at 2:57
  • $\begingroup$ @ajotatxe I guess this would be the form it would take on. Yes. $\endgroup$ – Zach L Jun 26 '15 at 2:58

From $$\frac{f'(x)}{g'(x)}=\frac{g(x)}{f(x)}$$ we get $$2f(x)f'(x)=2g(x)g'(x)$$ and integrating $$f(x)^2=g(x)^2+C$$ or $$f(x)=\pm\sqrt{g(x)^2+C}$$ provided that $C\ge-g(x)^2$ for every $x$ in the domain of $g$.

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    $\begingroup$ This condition on $C$ could also be weakened it seems; assuming it holds in a neighborhood of where the limit is being taken (in the above case, for all large enough $x$) would be enough. $\endgroup$ – Titus Jun 26 '15 at 3:08
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    $\begingroup$ great example of: if you write down what you mean explicitly, the answer is almost immediate $\endgroup$ – MichaelChirico Jun 26 '15 at 4:19

Here is a list of limits related to the OP question. L'Hospital's rule cannot be used for each.

  1. $\lim\limits_{x\rightarrow 0}\dfrac{x^{2}\sin \left( \frac{1}{x}% \right) }{\sin x}$

  2. $\lim\limits_{x\rightarrow \infty }\dfrac{x-\sin x}{x+\sin x}$

  3. $\lim\limits_{x\rightarrow \infty }\dfrac{\left( 2x+\sin 2x\right) }{% \left( 2x\sin x\right) e^{\sin x}}$

  4. $\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{2+x^{2}}}{x}$

  5. $\lim\limits_{x\rightarrow 0}\dfrac{1-\cos x}{\cos x}$

  6. $\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{9x+1}}{\sqrt{x+1}}$

  7. $\lim\limits_{x\rightarrow 0}\dfrac{\sqrt{x}}{\sqrt{\sin x}}$

  8. $\lim\limits_{x\rightarrow 0}\dfrac{\cot x}{\csc x}$

  9. $\lim\limits_{x\rightarrow \left( \pi /2\right) ^{-}}\dfrac{\sec x}{% \tan x}.$

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    $\begingroup$ $\lim_{x\rightarrow 0} (1-\cos(x))/\cos(x)$ is just $0/1$ $\endgroup$ – chi Jun 26 '15 at 8:54

You're basically asking if we can find functions $f, g$ such that:

$$\frac{f'}{g'} = \frac{g}{f}$$ i.e.

such that $$ff'=gg'$$

And in this particular case, you have $ff'=x$, of which the solutions are of the form $x \mapsto \sqrt{x^2 + c}$ (you can see that $x \mapsto x$, $x>0$ is a particular case when $c=0$).

Then you have found the solutions for $ff' = nx^{2n-1}$.

Now for any function $v$, you can look for solutions of the equation:

$$ff' = v$$ and if you find two solutions $f, g$ that don't touch $0$, the same phenomenon will occur with their ratio.

EDIT: We could probably look for bigger cycles, like

$$\frac{f'}{g'}=\frac{u}{v} \text{, and } \frac{u'}{v'} = \frac{g}{f}$$

But it looks a little bit more difficult to study, as it feels like a lot of things can happen.

  • $\begingroup$ Don't I have the solutions where ff' = x? $\endgroup$ – Zach L Jun 26 '15 at 3:16
  • $\begingroup$ @ZachL true and the other one is wrong too! edit: corrected $\endgroup$ – hHhh Jun 26 '15 at 3:18
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    $\begingroup$ would be fun to see a longer cycle, now that you mention it! anyone? $\endgroup$ – MichaelChirico Jun 26 '15 at 4:20

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