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I want to find this limit:

$$\lim_{h \to 0}\frac{(x+h)^3-x^3}{h}$$

I know that I can expand $(x+h)^3$ and with a little of algebra, the limit is $3x^2$.

My question: Can I use L'Hôpital's rule to find this limit?, if I use L'Hôpital's I have to differentiate numerator and denominator, but the derivative of $x^3$ is the limit that I try to find. Using L'Hôpital's here is a circular reasoning?

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    $\begingroup$ If L'Hopital's rule is allowed in your toolbox, then you can use it. If you are trying to prove the derivative of $x^3$ is $3x^2$, then it is circular reasoning. $\endgroup$ – Stephen Montgomery-Smith Nov 27 '13 at 1:09
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    $\begingroup$ Yes, it is circular reasoning, regardless of whether l'Hôpital's rule is one of the standard tools you can use. More interestingly, this illustrates that many uses of l'Hôpital's rule one encounters are actually superfluous: If $f'(a)$ and $g'(a)$ exist, and $g'(a)\ne0$, then $\displaystyle \lim_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(a)}{g'(a)}$, because $\displaystyle \lim_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}= \frac{\lim_{x\to a}\frac{f(x)-f(a)}{x-a}}{\lim_{x\to a}\frac{g(x)-g(a)}{x-a}}$. $\endgroup$ – Andrés E. Caicedo Nov 27 '13 at 1:19
  • $\begingroup$ I have to agree with Andres Caicedo. It is impossible to apply L'Hospital without finding derivative. And this limit is the process through the derivative is being calculated. In my view L'Hospital should not be used for 1st order limits. By 1st order limit I mean limits involving only $o(x)$ when $x \to 0$. It is better to use L'Hospital in cases of $o(x^{2}), o(x^{3})$ etc. $\endgroup$ – Paramanand Singh Nov 27 '13 at 4:50
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    $\begingroup$ Whether or not it is circular to approach this problem with L'Hopital's Rule really depends on the context of the question: are we trying to prove ${d\over dx}(x^3)=3x^2$ from first principles here, or for example, is this a final exam problem where one is free to appeal to any and all methods learned in the course? The former is circular while the latter, IMO, is not. $\endgroup$ – JohnD Nov 27 '13 at 5:49
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    $\begingroup$ The question is: Do you "know" the derivative of x^3 or not? If you "know" it you can just say "that expression is just the definition of f'(0) where f(x) = x^3, and the derivative of x^3 is 3x^2, so the limit is 3x^2". Of course using l'Hopital's rule is not wrong. If you had a list of 100 problems that can be solved all using l'Hopital and this one was on the list, it would be quite natural to apply l'Hopital without much thinking. $\endgroup$ – gnasher729 Apr 28 '14 at 14:23
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Since the limit is of type ${0\over 0}$ with respect to "limiting variable" $h$ (rather than $x$), the derivative in L'Hopital's Rule should be taken with respect to $h$:

$$\lim_{h\to 0}{(x+h)^3-x^3\over h}=\lim_{h\to 0}{{d\over dh}((x+h)^3-x^3)\over {d\over dh}(h)}=\lim_{h\to 0} {3(x+h)^2\over 1}=3x^2$$

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JohnD's very nice answer generalizes easily to higher powers:

$$\lim_{h\to 0}{(x+h)^n-x^n\over h} =\lim_{h\to 0}{{d\over dh}((x+h)^n-x^n)\over {d\over dh}(h)} =\lim_{h\to 0} {n(x+h)^{n-1}\over 1} =nx^{n-1} $$

Of course this uses the derivative of the function you are trying to differentiate to find the derivative of the function you are trying to differentiate, so it might be circular.

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