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I still do not know the answer, so I am looking for hints and tips here. Note that I want to do this with substitution, or at least not by taking derivatives.

Evaluate: (that's a cube root on the numerator)

$$\lim _{x \rightarrow 1}{\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}}$$

The example in the book suggests substituting to make the problem simpler, but I tried using $t = \sqrt{x}-1$ and the numerator also, but I still get a divide by zero in either case.

How do you make an educated guess on what to use for $t$ and substitute?

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Is this calculus? Have you learned L'Hopital's Rule yet? – abiessu Aug 12 '13 at 21:22
@abiessu It doesn't seem likely that he has learned about L'Hôpital yet. – Harald Hanche-Olsen Aug 12 '13 at 21:24
You may try too $x=1+t$ and use $(1+t)^a-1 = at+O(t^2)$. – Raymond Manzoni Aug 12 '13 at 21:26
up vote 6 down vote accepted

Well, you want a substitution so that you can remove everything in terms of $x$ and convert everything to being in terms of $t$. I recommend $t=\sqrt[6]{x},$ personally, since both $\sqrt[3]{x}$ and $\sqrt{x}$ can be expressed as powers of $\sqrt[6]{x},$ for non-negative $x$.

Once you've done that, the difference of squares and difference of cubes formulas should come in handy.

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@Leonardo: If we've got a function of the form $$f(x)=\frac{p(x)}{q(x)},$$ where $p(x)$ and $q(x)$ are monic polynomials with $q(x)$ a non-constant polynomial, then we can divide the leading terms' exponents to find $\lim_{x\to\infty}f(x)$. This example shows that we can do this in other situations, too. Ultimately, both sorts of situations can be resolved using l'Hôpital's rule, which you've apparently not yet encountered. – Cameron Buie Aug 12 '13 at 21:50

Expanding on Cameron's answer a bit, we have:

$$\frac {t^2 - 1} {t^3 - 1} = \frac {(t -1)(t+1)} {(t-1)(t^2 + t + 1)},$$ so $$\lim_{t\to 1} \frac {t^2 - 1} {t^3 - 1} = 2/3.$$

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This is strange, I thought I could only do that if $t$ approaches infinity. I could have divided the exponents in the original form without substitution and the answer is indeed $2/3$ but I thought it was the wrong method. – Leonardo Aug 12 '13 at 21:32
You can cancel the factors of $(t-1)$ so long as $t\neq 1$, and recall that the point $t=1$ does not enter into the calculation of the limit at 1. – Eric Auld Aug 12 '13 at 21:40
Very interesting, thank you for elaborating that. – Leonardo Aug 12 '13 at 21:44

My approach, if I were to do it by algebra alone, would be to use the two identities $u^2-1=(u-1)(u+1)$ and $u^3-1=(u-1)(u^2+u+1)$ with $u=\sqrt{x}$ and $u=\sqrt[3]{x}$, respectively. So you get $$\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1},$$ from which the limit is easily computed.

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$$ \lim_{x \to 1}{\sqrt[3]{x} - 1 \over \sqrt{x} - 1} = \lim_{x \to 1}{\left(1/3\right)\,x^{-2/3} \over \left(1/2\right)\,x^{-1/2}} = {2 \over 3} $$

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