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Does L'Hopitals Rule hold for second derivative, third derivative, etc...? Assuming the function is differential up to the $k$th derivative, is the following true?

$$\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)} = \lim_{x\to c} \frac{f''(x)}{g''(x)} = \cdots = \lim_{x\to c} \frac{f^{(k)}(x)}{g^{(k)}(x)} = \lim_{x\to c} \frac{f^{(k+1)}(x)}{g^{(k+1)}(x)}$$

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    If $f'$ and $g'$, etc. satisfy the hypotheses then yes. – 2017-02-09

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When L'Hopital's rule is stated by saying that $\displaystyle \lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)},$ one of the hypotheses is that $\lim_{x\to c}f(x) = \lim_{x\to c} g(x) = 0,$ or else both limits are among the two objects $\pm\infty.$ In order to take it one more step and say $\displaystyle \lim_{x\to c} \frac{f'(x)}{g'(x)} = \lim_{x\to c}\frac{f''(x)}{g''(x)},$ you would need to know that $\lim_{x\to c} f'(x) = \lim_{x\to c} g'(x) = 0$ or else both of those are among $\pm\infty.$ And you don't need to know anything beyond L'Hopital's rule as stated at the outset in order to know that.

Another hypothesis of L'Hopital's rule is that the limit to the right of the "equals" sign exists. You have no guarantee that L'Hopital's rule can be applied until you know that.

(Where you wrote $n\to c,$ I wrote $x\to c$.)

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    Actually, in the case the $g\to \infty$, then LRH is still applicable, even if $\lim f$ fails to exist. – 2017-02-09
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    @Dr.MV :-/ That's that weird case I'd personally just ignore. Why would you even apply L'Hospital's rule in that case? – 2017-02-09
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    @simplybeautifulart In the case for which $f$ approaches a finite limit, it's applicable but unnecessary. But if the numerator does not approach a finite limit, then LHR is applicable and might be useful. – 2017-02-09
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    @Dr.MV Well that is almost too obvious. :-| – 2017-02-09
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    @Dr.MV : Can you cite an instance of the application of the rule in that situation? – 2017-02-09
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    There are a lot applicable examples that might be considered trivial. Here is one that might be less obvious. Let $f(x)=\log(1/x)+\int_0^1 \frac{1-t^x}{1-t}\,dt$ and $g(x)=x$. Without knowing the limit of $f(x)$ as $x\to \infty$ (it's $\gamma$) we have $$\lim_{x\to \infty}\frac{f(x)}{g(x)}=\lim_{x\to \infty}\left(-\frac1x-\int_0^1 \frac{\log(t)t^x}{1-t}\,dt\right)=0$$ – 2017-02-09
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    I think the comment of Dr. MV only simplifies the application of L'Hospital for $\infty/\infty$ form because we don't bother to check the behavior of numerator. For example suppose $f'(x) \to L$ as $x \to \infty$ and we need to show that $f(x)/x \to L$. Then without worrying about behavior of $f$ we can apply L'Hospital here. BTW the worry for behavior of $f$ in this case is not too much of a load for experienced people, but perhaps may be significant overhead for a beginner. – 2017-02-09
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    @Dr.MV: It appears that you gave your reply when I was typing my comment. – 2017-02-09
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    @ParamanandSingh Yes. But there are cases for which the application might be useful. I constructed the case I gave in the comment with knowledge of the asymptotic expansion of $f(x)$. And with that knowledge, one wouldn't even need LHR. But if one was unaware, then a quick application of LHR does the trick. Of course one would need to justify differentiating under the integral and interchanging the limit and the integral (the DCT does the trick). – 2017-02-09
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    BTW +1 for the line containing "no guarantee". Beginners are perhaps so much fond of this rule that they overlook this aspect. – 2017-02-09
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Well, let's see. Under the specified conditions and $f(c)=g(c)=0$ (or both go to $\pm\infty$), then

$$\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$$

Now, if $f'(x)$ and $g'(x)$ still hold all those conditions and $f'(c)=g'(c)=0$ (or both go to $\pm\infty$), then

$$\lim_{x\to c}\frac{f'(x)}{g'(x)}=\lim_{x\to c}\frac{f''(x)}{g''(x)}$$

And you may repeat this process until the limit is easily evaluated or it fails to uphold the conditions of L'Hospital's rule.