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I'm trying to solve the following limit (without differentiation):

$$\lim \limits_{h \to 0}{\frac{\sqrt[3]{x+h} - \sqrt[3]{x}}{h}}$$

I know that multiply by the conjugates can be helpful when the roots are square. However, that just makes it messy when I do it with cube roots. So, I decided to modify the conjugate as $(x+h)^{2/3} + x^{2/3}$, in hopes that it'll make things easier. However, I ended up with a messier fraction than I had begun with.

Hints are welcome. (No solution please, I just need a kickstart.)

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    Just draw a graph of $f(x)=\sqrt[3]{x}$ in a neighbourhood of the origin. Hint: $f(x)$ is the inverse function of $x^3$. – 2017-02-10
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    Duplicate: [Finding derivative of $\sqrt[3]{x}$ using only limits](https://math.stackexchange.com/q/112865/201168). (*Found using [Approach0.xyz](https://approach0.xyz/search/?q=%24\lim_{h%20\to%200}{\frac{\sqrt[3]{%28x%2B%E2%80%8C%E2%80%8Bh%29}%20-%20\sqrt[3]{x}}{h}}%24&p=1)*) – 2017-02-10
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    @Workaholic -- Oh, oops. Should I delete this question? – 2017-02-10
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    @SirJony No, just flag it as a duplicate. – 2017-02-11

3 Answers 3

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Since $a^3-b^3 = (a-b)(a^2+ab+b^2) $, $a-b =\dfrac{a^3-b^3}{a^2+ab+b^2} $ so $a^{1/3}-b^{1/3} =\dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}} $.

Note that this works for $a^{1/n}$ for any integer $n \ge 2$.

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You're trying to use some identity like $(a-b)(a^2+b^2)=?$.

Instead, try an identity like $(a-b)(a^2+ab+b^2)=?$.

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Hint...what is the definition of the derivative of the function $y=x^{\frac 13}$?

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    Yes, I saw that, too. However, I'm technically not supposed to know derivatives, yet. So, I'm trying to use other methods. :) +1 Anyways. – 2017-02-10
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    OK, so at least you know what the answer is supposed to be. Maybe you should edit the question to say without using differentiation from first principles. Or is this in fact meant to be an exercise in differentiation from first principles? – 2017-02-10
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    No, the exercise book (Demidovich) I'm going by hasn't mentioned derivatives yet. So, this is purely limits right now. I've updated the question to match your request. – 2017-02-10