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$$ \lim_{x\to 1} \frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1} $$

I multiply it with:

$$ \frac{ \sqrt{x} + 1}{\sqrt{x} + 1} $$

And I get :

$$ \lim_{x\to 1} \frac{ \sqrt{x}^2 - 1^2}{(\sqrt[3]{x} - 1) * (\sqrt{x} + 1)} $$

But the solution is still division by 0 and not possible, so I think I've made a mistake somewhere..

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From that step, you can apply L'Hopital's rule. – Ian Coley Dec 16 '13 at 15:15
@mathlove sorry about that, typo while writing... fixed. :) – user95523 Dec 16 '13 at 15:16
@IanColey What exactly is that rule? – user95523 Dec 16 '13 at 15:17
@IanColey I did apply it and I get a weird result (1/2 * x^(-1/2) -1) / (1/3 * x^(-2/3) - 1) – user95523 Dec 16 '13 at 15:22

5 Answers 5

up vote 10 down vote accepted

Hint $\ $ Let $\ X = x^{1/6}.\ $ Then it is $\ \dfrac{X^3-1}{X^2-1} = \dfrac{X^2+X+1}{X+1},\ $ no longer indeterminate.

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+1 so smooth... – Avitus Dec 16 '13 at 15:31
Looks like a good answer to me, compared to others.. however to solve problems in this way, you need way too much experience and knowledge in math. – user95523 Dec 16 '13 at 15:33
+1 for being nice and elegant. @user95523 This is more of a simple substitution, you don't need too much experience to use it. – Link Dec 16 '13 at 15:43
@Link you're correct, I don't need much experience to use it, but I wouldn't have thought of it myself.. usage is very easy to understand :) – user95523 Dec 16 '13 at 15:52
@user95523 Well in all honesty, neither did I. But I guess that's how math is. – Link Dec 16 '13 at 16:07

You basically want to insert $1$ written as a “conjugate” that removes the radical divided by itself. For $\sqrt{x}-1$ it is $\sqrt{x}+1$, for $\sqrt[3]{x}-1$ it is $\sqrt[3]{x^2}+\sqrt[3]{x}+1$. So you get \begin{align} \lim_{x\to1}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}&= \lim_{x\to1}\left((\sqrt{x}-1)\frac{\sqrt{x}+1}{\sqrt{x}+1}\right) \left(\frac{1}{\sqrt[3]{x}-1} \frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\right) \\[2ex] &=\frac{x-1}{\sqrt{x}+1}\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1}{x-1} \end{align} and the indetermination goes away.

It's the same as Bill Dubuque's answer, actually, but without substitution.

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Here is a version computationally close to L'Hospital's Rule, though it does not mention him. Note that $$\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}= \frac{\frac{\sqrt{x}-1}{x-1} }{\frac{\sqrt[3]{x}-1}{x-1} }.\tag{1}$$

If we look at the expression on the right of (1), we see that the limit as $x\to 1$ of the top is by definition the derivative of $\sqrt{x}$ at $x=1$. Similarly, the limit of the bottom is the derivative of $\sqrt[3]{x}$ at $x=1$. Compute these derivatives, using the ordinary "Power Rule."

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Hint: $$\frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1} =\frac{3}{2} \cdot \frac{ e^\frac{\ln x}{2} - 1}{\frac{\ln x}{2}} \cdot \left( \frac{ e^\frac{\ln x}{3} - 1}{\frac{\ln x}{3}}\right)^{-1} $$

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Wait, what? Where did that come from, sir? – user95523 Dec 16 '13 at 15:23
Majority of people underestimate power of well known limits and tries to bash limit with L'Hospital Rule instead of sit and think for a while. I'm sure you know that $\lim \frac{e^x-1}{x} = 1$ as $x$ goes to $0$ and following trick: $\sqrt{x} = x ^ {\frac{1}{2}}= e^{\frac{\ln x}{2}}$ – Stephen Dedalus Dec 16 '13 at 15:29
Yes, I'm aware sqrt(x) = x^(1/2) but I dont know how do we get the other one e^(lnx/2) – user95523 Dec 16 '13 at 15:31
Emm $e^{\ln x} = x$ – Stephen Dedalus Dec 16 '13 at 15:34

Hint: You are looking to apply L'Hopital's Rule. This rule basically states the following:

$$\lim_{x\to 1} \frac{f(x)}{g(x)} = \lim_{x\to 1} \frac{f'(x)}{g'(x)}$$

Note: you can keep applying this rule multiple times, so it can go on and on until you get the value of the limit.

So keep calculating the limit of the derivative until you get one that is not division by zero or undefined, that value is the answer is the limit of your original question.

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Yes, I've tried using that rule, but the result ends up so weird with many fractions and hard to continue. – user95523 Dec 16 '13 at 15:29
@user95523 did you try it more than once? i.e. Do it on your result again. – Link Dec 16 '13 at 15:31
Yes, I've tried doing it multiple times, I keep getting fractions over and over.. – user95523 Dec 16 '13 at 15:32

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