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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.

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  • $\begingroup$ It's a little messy, but you do need to plug the expansion into itself. What you need to understand is that you only need to compute terms to $x^5$, so it is not as horrific as it looks. $\endgroup$
    – Ron Gordon
    Jan 16, 2016 at 17:23

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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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You have to use the Taylor's series for the term $\sin(\sin(x))- x(1-x^2)^{\frac 13}$, so that to develop this term up to 5th order. For example: $x(1-x^2)^{\frac 13}= x(1- \frac 13 x^2+ \frac 19 x^4 +O(x^5)) $. On the other hand, for the term $\sin(\sin(x))$ use Taylor for the composition of two functions.

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