Compute $\lim\limits_{x \to 0} \frac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$ 
Compute $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$

Original question was to solve $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+ x } - \sqrt[3]{1-x }}{x}$ and it was solved by adding and subtracting 1 in denominator.  Making it in the form $\lim\limits_{x \to a}\dfrac {x^n - a^n}{x-a} = ax^{n-1}$
How to solve for above limit without using lhopitals rule?

If you already know how to find the limit $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+ x } - \sqrt[3]{1-x }}{x}$ and also that $\dfrac{\sin x}x$ tends to 1, then you can simply use
$$
\frac {\sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x}}{x} = 
\frac {\sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x}}{\sin x} \cdot \frac{\sin x}x.
$$
