Factoring limits

Question

Couldn't find this limit someone help me?

$$\lim_{x\rightarrow0} \frac{ (1+x)^{1/3} - (1-x)^{1/3}}{x}$$

I tried to take $x^{1/3}$ common from above expression

Answer 1

I suppose that you mean$$\lim_{x\to0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}x.$$If so, this is just$$\lim_{x\to0}\frac{f(x)-1}x-\lim_{x\to0}\frac{g(x)-1}x\text,$$with $f(x)=\sqrt[3]{1+x}$ and $g(x)=\sqrt[3]{1-x}$. So, your limit is $f'(0)-g'(0)=\frac13+\frac13=\frac23$.

Answer 2

Let $A=(1+x)^{1/3}$ and $B=(1-x)^{1/3}$ then $$2x=A^3-B^3=(A-B)(A^2+AB+B^2)$$ Hence, as $x\to 0$, we have that $A\to 1$, $B\to 1$ and $$\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}x=\frac{A-B}{x}=\frac{2}{A^2+AB+B^2}\to\frac{2}{3}.$$

Answer 3

Another approach using a definition of the derivative i.e. that

$$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$$

Applying that in the second step:

$$ \begin{align} \lim_{x\to 0} \frac{ (1+x)^{1/3} - (1-x)^{1/3}}{x} &= 2\lim_{x\to 0} \frac{ (1+x)^{1/3} - (1-x)^{1/3}}{2x} \\ &= 2 \frac{d}{dx}(x \mapsto x^{1/3})|_{x=1} \\&= 2 \frac{1}{3}x^{-2/3}|_{x=1} \\&= \frac{2}{3} \end{align}$$