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?

Just another way using Taylor series.
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ Now, using the generalized binomial theorem 
$$\sqrt[3]{1+\sin(x)}=1+\frac{x}{3}-\frac{x^2}{9}+\frac{x^3}{162}+O\left(x^4\right)$$
$$\sqrt[3]{1-\sin(x)}=1-\frac{x}{3}-\frac{x^2}{9}-\frac{x^3}{162}+O\left(x^4\right)$$
Using the above,
$$\sqrt[3]{1+\sin(x)}-\sqrt[3]{1-\sin(x)}=\frac{2 x}{3}+\frac{x^3}{81}+O\left(x^4\right)$$
$$\dfrac{ \sqrt[3]{1+\sin (x)} - \sqrt[3]{1- \sin (x) }}{x}=\frac{2}{3}+\frac{x^2}{81}+O\left(x^3\right)$$ which shows the limit and also how it is approached.
Making the problem more general considering 
$$A=\frac{(1+\sin (x))^k-(1-\sin (x))^k}{x}$$ and using the same process, we should obtain
$$A=2k+\frac{k \left(k^2-3 k+1\right)}{3} x^2+O\left(x^3\right)$$
