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Can this limit be evaluated without l'hopital's rule?


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sure, the 3rd square root is differentiable in 8. And your expression is (assuming $h\to 0$) just a difference quotient – Quickbeam2k1 Mar 7 '14 at 15:57
It probably should be $h \to 0$ in the limit. – user133281 Mar 7 '14 at 15:59
@Rogers mixedmath's solution is actually the simplest and most intuitive. Try to work it out based off the hint. – Tyler Holden Mar 7 '14 at 16:10
l'Hopital's rule can actually be derived from other rules; it is not an "axiom." So if a limit can be solved with l'Hopital's rule, it can also be solved without l'Hopital's rule. – TMM Mar 7 '14 at 16:29
@BenMillwood Agh! Sorry, made a typo, comment now corrected. – Andrea L. Mar 7 '14 at 20:36
up vote 11 down vote accepted

$$x^3-y^3=(x-y)(x^2+xy+y^2)\implies x-y=\frac{x^3-y^3}{x^2+xy+y2}$$

Now we put

$$x=\sqrt[3]{8+h}\;,\;\;y=2\implies \sqrt[3]{8+h}-2=\frac{8+h-8}{(8+y)^{2/3}+2\sqrt[3]{8+h}+4}\implies$$

$$\frac{\sqrt[3]{8+h}-2}h=\frac1{(8+h)^{2/3}+2\sqrt[3]{8+h}+4}\xrightarrow[h\to 0]{}\frac1{8^{2/3}+2\sqrt[3]8+4}=\ldots$$

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Some net lag I just had, I didn't see your answer or Yiorgos' before I posted. No sense in having 3 identical answers. (+1) – robjohn Mar 7 '14 at 16:27

Try to let: $u= (8 + h)^{1/3} $

Thus $u^3 = 8 + h$

$h = u^3 - 8$

Then the limit becomes:

$$\lim_{u\to2}\frac{u-2}{u^3-8} = \lim_{u\to2}\frac{u-2}{(u-2)(u^2+2u+4)}=\lim_{u\to2}\frac{1}{u^2+2u+4}$$

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This seems to be the same basic idea as DonAntonio, but it is expressed much more simply and clearly. +1 – jpmc26 Mar 8 '14 at 0:34
Haha wow, I don't go on the computer for 2 days and I see this. Thanks everyone! :) – user133458 Mar 10 '14 at 1:55
This indeed is the same idea as in my answer but seriously simpler and nicer. I can't understand how come my answer was accepted and not this one. Anyway, +1 – DonAntonio Mar 11 '14 at 18:47


Think about the definition of a derivative, recognize this expression as the derivative of something. (And yes, this can be found without l'Hopital's rule).

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One neat way to solve some limits is to use Taylor series. If you recall the Maclaurin expansion $$(1 + x)^n = 1 + nx + \mathcal{O}(x^2)$$ you can see that $$\sqrt[3]{8 + h} -2 = 2\sqrt[3]{1 + h/8} -2 \approx 2\left(1 + \frac{h}{24}\right) -2 = \frac{h}{12}$$ so the limit becomes $$\lim_{h \to 0} \frac{h/12}{h} = \frac{1}{12}.$$

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It's a neat way to solve a lot of limits, I have never touched L'Hopital & co. since I knew Taylor expansions and small-oh notation. – Matteo Italia Mar 8 '14 at 1:28
@MatteoItalia it's the same for me. – giordano Mar 8 '14 at 10:06

\begin{align} \frac{\sqrt[3]{8+h}-2}{h}&= \frac{\sqrt[3]{(8+h)^2}+2\sqrt[3]{8+h}+4}{\sqrt[3]{(8+h)^2}+2\sqrt[3]{8+h}+4} \cdot \frac{\sqrt[3]{8+h}-2}{h} \\&=\frac{(8+h)-8}{h} \cdot\frac{1}{\sqrt[3]{(8+h)^2}+2\sqrt[3]{8+h}+4} \to\frac{1}{12}. \end{align}

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