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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$\begingroup$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 $x\to c$.)
$\endgroup$ 10 $\begingroup$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.
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