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I can't find the limit of the the following:

$$\lim_{p\to1} \frac{ p^{1/3} - 1 }{p - 1}$$

Any ideas?

Alfonso Fernandez
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Luke Taylor
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3 Answers3

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Putting $p^{\frac13}=q\implies p=q^3$ as $p\to1, q\to1$

$$\lim_{p\to1}\frac{p^{\frac13}-1}{p-1}$$

$$=\lim_{q\to1}\frac{q-1}{q^3-1}$$

$$=\lim_{q\to1}\frac{(q-1)}{(q-1)(q^2+q+1)}$$

$$=\lim_{q\to1}\frac1{q^2+q+1}\text{ as }q-1\ne0\iff q\ne1\text{ as } q\to1$$

$$=\cdots$$

lab bhattacharjee
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Besides the nice answer you already got, what about l'Hospital?:

$$\lim_{p\to 1}\frac{p^{1/3}-1}{p-1}=\lim_{p\to 1}\;\frac13p^{-2/3}=\frac13$$

DonAntonio
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  • Thank you. Looks complicated though. :P – Luke Taylor Nov 16 '13 at 14:52
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    It really isn't and, in fact, l'Hospital is one of the strongest methods to calculate some particular kinds of limits...so much that some limits seem to be almost impossible to evaluate otherwise. – DonAntonio Nov 16 '13 at 14:55
  • Okay, thank you, I'll check out l'Hospital for future reference. :) – Luke Taylor Nov 16 '13 at 14:59
  • L'Hôpital's rule is powerful, but it's usually pedagogically wrong at the level of this question. Understanding why L'Hôpital works requires you to be familiar with derivatives, differentiability, the sandwich theorem and Cauchy's mean value theorem just to get started. In an elementry calculus test, be forewarned, I know professors that will mark an answer using L'Hôpital as wrong, unless the student also provides definitions and proofs for all of the above if they haven't been presented in class yet. – Euro Micelli Nov 16 '13 at 15:23
  • Any instructor, teacher and/or proffesor I know, up and including me, will downmark any question that isn't well explained and justified, and this of course includes stuff that wasn't covered in the course. We're assuming here that l'H was already taught, otherwise it doesn't fit. – DonAntonio Nov 16 '13 at 15:29
  • However in this question it is intrinsically not possible to use L'Hospital because of circular reasoning. The given limit is actually the derivative of $f(x) = x^{1/3}$ at point $1$ and and L'Hospital assumes the use of differentiation. The limit therefore has to be justified by the use of algebraic manipulations. – Paramanand Singh Nov 16 '13 at 15:56
  • Well, that's one possibility, yet I could simply know derivatives and l'H, and I could also know that $$f'(1):=\lim_{h\to 0}\frac{(1+h)^{1/3} -1}h$$which, of course, is equivalent but not the same as the given limit. I can't see the circularity here if derivatives and l'H are already known. – DonAntonio Nov 16 '13 at 15:58
  • To DonAntoio: The derivative can't be known unless you calculate this limit by some other means. Derivatives are not magical formulas but they are result of limit operations (which are evaluated by techniques other than differentiation). That's why I see them them as circular. In this case if you L'Hospital then you have to use the fact that $(x^{1/3})' = (1/3)x^{-2/3}$ and this means you already know that $\lim_{x \to a}(x^{1/3} - a^{1/3})/(x - a) = (1/3)a^{-2/3}$ and then you can directly put $ a = 1$ to get answer. – Paramanand Singh Nov 16 '13 at 16:03
  • In my opinion L'Hospital can't be used to evaluate a limit of the form $\lim_{x \to a}(f(x) - f(a))/(x - a)$. – Paramanand Singh Nov 16 '13 at 16:04
  • There are lots of methods to calculate derivatives after one knows the definition by means of limits, @ParamanandSingh...fortunately enough. One of them is by means of the derivative of the inverse function. So I could know the very easy (by definition or whatever) $$(x^3)'=3x^2\implies y=x^{1/3}\implies x=y^3\implies y'=\frac1{(y^{-1})'}=\frac1{3x^2}=\frac1{3y^{3/2}}\ldots$$ and thus I could know the derivative not by means of the definition... – DonAntonio Nov 16 '13 at 16:07
  • Imo l'H can be used to evaluate a limit of the kind you wrote in your last comment @ParamanandSingh as long as the derivatives and the method are already known. I agree that with basic stuff that can be weird from an educative point of view, yet when one reaches l'H and stuff one has already studied the basics enough to understand what's going on. – DonAntonio Nov 16 '13 at 16:10
  • Still I would prefer that L'Hospital and its even more powerful cousin Taylor series should be used for things which are really tough and not for the particular simple problem here. A problem suitable for use of L'Hospital is at http://math.stackexchange.com/questions/569043/a-limit-problem-related-to-log-sec-x Unfortunately I am yet to solve this one. – Paramanand Singh Nov 16 '13 at 17:18
  • As for making some mental muscle I agree with you in that, @ParamanandSingh: leave stronger weapons for stronger problems. – DonAntonio Nov 16 '13 at 17:19
  • let us continue this discussion in chat – Paramanand Singh Nov 16 '13 at 17:20
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Hint

Let $$f(p)=p^{1/3}$$ Can you see why your limit is $f'(1)$?