# Find limit of fraction without derivatives

How to find limit of this?

$$\lim_{h \rightarrow 0} \frac {\sqrt[3]{x+h} - \sqrt[3]{x}} {h}.$$

I see here the derivative definition, so I understand in "derivative mind" that answer has to be $$\frac{1}{3} x^{-\frac{2}{3}},$$but in the book this expression goes before derivatives, so it must be another approach to find this limit.

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Rationalize the numerator. –  David H Jul 26 '14 at 11:08

Hint: $$(a-b)(a^2+ab+b^2)=a^3-b^3.$$

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Rationalise the numerator by the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$

$\lim _{h\rightarrow 0}{\frac {\sqrt [3]{x+h}-\sqrt [3]{x}}{h}}$

$=\lim _{h\rightarrow 0}{\frac {\sqrt [3]{x+h}-\sqrt [3]{x}}{h}}{\frac { \left( x+h \right) ^{2/3}+\sqrt [3]{x+h}\sqrt [3]{x}+{x}^{2/3 }}{ \left( x+h \right) ^{2/3}+\sqrt [3]{x+h}\sqrt [3]{x}+{x}^{2/3}}}$

$=\lim _{h\rightarrow 0}{\frac {(x+h)-x}{h \left( \left( x+h \right) ^{2/3}+ \sqrt [3]{x+h}\sqrt [3]{x}+{x}^{2/3} \right) }}$

$=\lim _{h\rightarrow 0}{\frac {1}{ \left( x+h \right) ^{2/3}+ \sqrt [3]{x+h}\sqrt [3]{x}+{x}^{2/3} }}$

$={\frac {1}{ \left( x \right) ^{2/3}+ \sqrt [3]{x}\sqrt [3]{x}+{x}^{2/3} }}$ by plugging in $h=0$

$={\frac{1}{3x^{2/3}}}$

=$\frac{1}{3}x^{-2/3}$

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Hint

Another one (assuming $x\neq 0$) :$$\sqrt[3]{x+h}=x^{\frac{1}{3}}\sqrt[3]{1+\frac{h}{x}}$$ Now, if you remember that, when $y$ is small, $(1+y)^a \simeq 1+ ay$, then now, replace $y$ by $\frac{h}{x}$ and so $$\sqrt[3]{x+h}=x^{\frac{1}{3}}\sqrt[3]{1+\frac{h}{x}}\simeq x^{\frac{1}{3}}(1+\frac{h}{3x})=x^{\frac{1}{3}}+\frac{h}{3x^{\frac{2}{3}}}$$

I am sure that you can take from here.

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Very useful for me. –  Leo Matyushkin Jul 26 '14 at 11:29
You are very welcome. Remember this approximation : you will use it a lot in your studies. Cheers :) –  Claude Leibovici Jul 26 '14 at 11:31

HINT:

Set $\displaystyle\sqrt[3]{x+h}=a\implies x+h=a^3,\sqrt[3]x=b\implies x=b^3; a^3-b^3=h$

As $\displaystyle h\to0, a\to b\implies a\ne b$

Finally, the limit should be $\displaystyle3/b^2=3/(x^{\dfrac23})$

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