Evaluate: $\lim\limits_{x \to 0} \dfrac{\sqrt[3]{\left(1+x\right)}-\sqrt[3]{\left(1-x\right)}}{x}$
$\begin{aligned}
\lim\limits_{x \to 0} \dfrac{\sqrt[3]{\left(1+x\right)}-\sqrt[3]{\left(1-x\right)}}{x} & = \lim\limits_{x \to 0} \dfrac{\left(1+x\right)^{1/3}-1 - \left(1-x\right)^{1/3}+1}{x} \\
& \\
& = \lim\limits_{x \to 0} \; \dfrac{\left(1+x\right)^{1/3}-1}{x} - \lim\limits_{x \to 0} \; \dfrac{\left(1-x\right)^{1/3}-1}{x} \\
& \\
& = L_1 - L_2
\end{aligned}$
Consider $L_1 = \lim\limits_{x \to 0} \dfrac{\left(1+x\right)^{1/3}-1}{x}$
Let $1+x=p$
Then as $x \to 0$, $p \to 1$
Also, $x = p-1$
$\therefore$ $L_1 = \lim\limits_{p \to 1} \dfrac{p^{1/3}-1}{p-1} = \dfrac{1}{3} \times 1^{\frac{1}{3}-1}=\dfrac{1}{3}$ $\left[\text{Note: }\lim\limits_{x \to a} \dfrac{x^n - a^n}{x-a}=na^{n-1}\right]$
Consider $L_2 = \lim\limits_{x \to 0} \dfrac{\left(1-x\right)^{1/3}-1}{x}$
Let $1-x=q$
Then, as $x \to 0$, $q \to 1$
Also, $x=1-q$
$\therefore$ $L_2 = \lim\limits_{q \to 1} \dfrac{q^{1/3}-1}{1-q} = - \lim\limits_{q \to 1} \dfrac{q^{1/3}-1}{q-1}=-\dfrac{1}{3} \times 1^{\frac{1}{3}-1} = - \dfrac{1}{3}$
$\therefore$ $\lim\limits_{x \to 0} \dfrac{\sqrt[3]{\left(1+x\right)}-\sqrt[3]{\left(1-x\right)}}{x} = L_1 - L_2 = \dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}$