# Limit in form of 0/0

If I multiply by I get zero/2x anyway. What manipulation needed to get 2/3?

• Check your multiplication again. Note you're dealing with cube roots and not square roots. Oct 15, 2013 at 15:34

You need to multiply by $$\left(\sqrt[3]{1+x}\right)^2+\left(\sqrt[3]{1+x}\right)\left(\sqrt[3]{1-x}\right)+\left(\sqrt[3]{1-x}\right)^2$$ on top and bottom, instead. This comes from the difference of cubes formula $$a^3-b^3=(a-b)(a^2+ab+b^2).$$ Had we been dealing with square roots instead of cube roots, then you would use the difference of squares formula, instead, as you are attempting to do implicitly.

Remember the identity $$a^3 - b^3 = (a-b)(a^2 + ab + b^2).$$ If you think of $a$ as $(1+x)^{1/3}$ and $b$ as $(1-x)^{1/3}$, you find that multiplying by $$\frac{(1+x)^{2/3} + (1+x)^{1/3}(1-x)^{1/3} + (1-x)^{2/3}}{(1+x)^{2/3} + (1+x)^{1/3}(1-x)^{1/3} + (1-x)^{2/3}}$$ will work.

$$\lim_{x\to 0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}=\lim_{x\to 0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}\cdot\frac{\sqrt[3]{(1+x)^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}}{x(\sqrt[3]{1-x^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}}=$$ $$=\lim_{x\to 0}\frac{1+x-1+x}{x(\sqrt[3]{1-x^2}+\sqrt[3]{(1-x^2}+\sqrt[3]{(1-x)^2}}=\lim_{x\to 0}\frac{2x}{x\left(\sqrt[3]{(1+x)^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}\right)}$$ $$=\lim_{x\to 0}\frac{2}{\sqrt[3]{(1+x)^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}}=\lim_{x\to 0}\frac{2}{\sqrt[3]{(1+0)}^2+\sqrt[3]{1-0^2}+\sqrt[3]{(1-0)^2}}=\frac{2}{3}$$

As $x\to0, |x|<1$

$\displaystyle(1+x)^{\frac13}=1+\frac13x+O(x^2)$ and $\displaystyle(1-x)^{\frac13}=1-\frac13x+O(x^2)$

$$\implies\lim_{x\to0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}x$$

$$=\lim_{x\to0}\frac{1+\frac13x+O(x^2)-\{1-\frac13x+O(x^2)\}}x$$ $$=\lim_{x\to0}\frac23+O(x)\left(\text{ cancelling }x\text{ as }x\ne0\text{ as }x\to0\right)$$

$$=\cdots$$