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$$\lim_{x \to 0} \frac{1}{x} + \frac{1}{x}$$ $$\lim_{x \to 0} \frac{2x}{x^2}$$

Since this evaluates to an indeterminate form $\frac{0}{0}$ we use l'Hospital's Rule:

$$\lim_{x \to 0} \frac{2}{2x}$$

Since this also evaluates to an indeterminate form $\frac{2}{0}$ we use l'Hospital's Rule again:

$$\lim_{x \to 0} \frac{0}{2}=0$$

I know that I could have simply divided both numerator and denominator by $x$ to get the same result. This is just an example to ask the question: Can l'Hospital's Rule be used recursively?

EDIT: Sorry, I messed up my example. I cannot think of a good example right now but the question still stands. Can one use l'Hospital's Rule recursively?

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    $\begingroup$ The form $\frac{2}{0}$ is not really indeterminate. It's $\pm \infty$ (which is still slightly indeterminate, but nowhere near as indeterminate as $\frac00$). So no L'Hospital. L'Hospital is for $\frac{0}{0}$ and $\frac{\pm\infty}{\infty}$, and even then there are cases where you need to be careful. $\endgroup$
    – Arthur
    Commented Sep 26, 2014 at 10:08
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    $\begingroup$ @Arthur: You lazy goose. You say absurdly brilliant things which you should expand and post as an answer but don't. $\endgroup$
    – Nick
    Commented Sep 26, 2014 at 10:47
  • $\begingroup$ $\lim_{x\to 0} \frac{1}{x}$ does not even exist! Here, Left Hand Limit $\neq$ Right Hand Limit. $\endgroup$
    – Nick
    Commented Sep 26, 2014 at 10:49

6 Answers 6

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L'Hôpital's Rule

Assuming that the following conditions are true:

  1. $f(x)$ and $g(x)$ must be differentiable
  2. $\frac{d}{dx}g(x)\neq 0$
  3. $\lim\limits_{x\to c} \frac{f(x)}{g(x)}= \frac{0}{0}\mbox{ or }\lim\limits_{x\to c} \frac{f(x)}{g(x)}= \frac{\pm\infty}{\pm\infty}$

Then, $$ \lim\limits_{x\to c} \frac{f(x)}{g(x)}= \lim\limits_{x\to c} \frac{\frac{d}{dx}f(x)}{\frac{d}{dx}g(x)}=L $$ Where $c$ and $L$ is any real number or $\pm\infty$.

So to answer your questions, yes, L'Hôpital's rule can be used repeatedly, provided that all of the above conditions are met. Since your example doesn't meet the aforementioned conditions, L'Hôpital's rule is not applicable.

Here is a case where L'Hôpital's rule is applicable multiple times, $$ \lim_{x\to \infty}\dfrac{e^x}{x^2}=\frac{\infty}{\infty} $$ Since the conditions are met, we can apply L'Hôpital's rule $$ \lim_{x\to \infty} \frac{\frac{d}{dx}e^x}{\frac{d}{dx}x^2}= \lim_{x\to \infty} \frac{e^x}{2x}= \frac{\infty}{\infty} $$ Notice that the conditions are met again, so now $$ \lim_{x\to \infty} \frac{\frac{d}{dx}e^x}{\frac{d}{dx}2x}= \lim_{x\to \infty} \frac{e^x}{2}= \infty $$ Therefore $$ \lim_{x\to \infty}\frac{e^x}{x^2}=\infty $$

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You're correct till the third step, although you can write $\dfrac1x+\dfrac1x = \dfrac2x$ instead of taking LCM.

However in the third step, $\lim\limits_{x\to0}\dfrac{2}{2x}$ is not an indeterminate form (not $\dfrac00$). Therefore, it's illegal inapplicable to use L'Hospital's Rule there.

Answer to 'recursive' question: I don't know what you mean by 'recursively', but it can be used repetatively.

Consider $$\lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2} = \lim\limits_{x\to0}\dfrac{\sin x}{2x} = \lim\limits_{x\to0}\dfrac{\cos x}{2} = \dfrac12$$ by using L'Hospital's Rule two times.

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    $\begingroup$ "Illegal"? I doubt any police officers are gonna be busting down the OP's door any time soon. $\endgroup$
    – beep-boop
    Commented Sep 26, 2014 at 15:52
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    $\begingroup$ I'm pretty much sure illegal is used in the context of math. Still, which word do you propose in lieu? $\endgroup$
    – taninamdar
    Commented Sep 26, 2014 at 15:54
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    $\begingroup$ @taninamdar, inapplicable? $\endgroup$
    – k170
    Commented Sep 26, 2014 at 16:01
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    $\begingroup$ It's not illegal. It's just wrong. The same as saying $1+1=3$ isn't warrant for arrest, but it's incorrect. $\endgroup$
    – beep-boop
    Commented Sep 26, 2014 at 17:03
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You can use the rule only if you are calculating $\lim \frac{f(x)}{g(x)}$ and $\lim f(x) = \lim g(x) = 0$ or $\lim f(x)=\lim g(x) = \infty$. The functions in $\frac2 {2x}$ do not satisfy this demand.

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$\lim_{x\to 0} \frac{2}{2x}$ is not an indeterminate form.

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    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$ Commented Sep 26, 2014 at 10:34
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    $\begingroup$ @TheGame: Part of the question (in the post title) was 'Is this a valid use of L'Hopital's rule?' $\endgroup$
    – paw88789
    Commented Sep 26, 2014 at 10:48
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    $\begingroup$ I posted that because I believed that given its length, this would be more appropriate as a comment. You might consider completing it by adding what the limit, or something like that :) $\endgroup$ Commented Sep 26, 2014 at 11:17
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    $\begingroup$ @TheGame: I would say that often very short answers can be quite illuminating. I figure if it makes the character minimum for a MSE answer, it's (potentially) long enough. :-) $\endgroup$
    – paw88789
    Commented Sep 26, 2014 at 12:23
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I see what you did there.

You tried to claim that: $\lim_{x\to 0} \dfrac {2}{x} = \lim_{x\to 0} \left(\dfrac {2}{x}\times\dfrac{x}{x}\right)$ and as that was an indeterminant form you could apply l'Hospital's rule.

l'Hopital's only works if you cannot remove indeterminacy by cancellation.   It doesn't work if you introduce indeterminacy through multiplication.

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Yeah, but make sure to work the correct way around when writing it out. L'Hopital's rule says $f'(x)/g'(x)$ converges and $f(x)/g(x)$ is an indeterminate form, then $f(x)/g(x)$ has the same limit. So if you want to iterate it say twice and take advantage of $f''(x)/g''(x)$ being nice, you need to first apply it it $f'(x)/g'(x)$ by showing $f'(x)/g'(x)$ is an indeterminate form and $f''(x)/g''(x)$ converges to deduce $f'(x)/g'(x)$ converges, then you have enough information to apply the rule to to $f(x)/g(x)$.

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