<s>[INST] Calculate $\lim_{x \to 0} \frac{\sin(\sin x)-x(1-x^2)^\frac{1}{3}}{x^5}$ by Taylor's theorem. I have to calculate : $$\lim_{x \to 0} \frac{\sin(\sin x)-x(1-x^2)^\frac{1}{3}}{x^5}$$ by using  Taylor's theorem. 
I know that :$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots$$
But I don't know how to continue.
Thanks.
 [/INST] Expand $\sin(\sin x)$ similar to $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$, i.e. $$\sin(\sin x)=\sin x-\frac{\sin^3x}3!+\frac{\sin^5 x}{5!}-...$$ near $x=0$.
And $(1-x^2)^{1/3}$ similar to $(1+x)^n=1+nx+\frac{(n-1)n}{2!}x^2+\frac{(n-2)(n-1)n}{3!}x^3+...$
Now, expand each $\sin x$ using taylor then forming combination of useful terms.
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