Limit in form of 0/0

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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

Using General Binomial Theorem

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

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

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

$$=\cdots$$