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Is there any type of limit that requires L'Hopital's rule to solve or can all limits be solved without using it?

If all limits can be solved without L'hopitals, is there some sort of proof or intuition for this?

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  • $\begingroup$ I think it's the consequence of some big theorems that not every limit can be solved, so in that sense there are some theorems that cannot be solved without using L'Hopital's. $\endgroup$ – T.J. Gaffney Dec 13 '14 at 21:49
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    $\begingroup$ As with all these things you can replicate the proof of l'Hôpital's rule for each specific example. The value of the rule is that it gives a general procedure, which is easy to use once you know it, and applicable across a range of common cases. Some instances have easy alternative proofs by other routes. For others where an alternative is not obviously available, why make life difficult? $\endgroup$ – Mark Bennet Dec 13 '14 at 21:51
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Usually Taylor's expansion has the same strength. However, if you prohibit both L'Hopital and Taylor, then I doubt that $\lim_{x\to 0} \frac{\sin x-x}{x^3}$ or $\lim_{x\to 0} \frac{\sin x-x+x^3/6}{x^5}$ can be solved elementary, without using the ideas of L'hopital or Taylor in some sense.

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If you cannot use L'Hopital's rule you can always use Cauchy's mean-value theorem together with the squeeze theorem.

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