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

Question

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.)

Answer 1

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 $n \ge 2$.

Answer 2

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

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

Answer 3

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