<s>[INST] Calculating $\lim \limits_{x \to 0} \frac{\cos x\sin x-x}{\sin^3 x}$ using taylor series I am given the following limit, and need to calculate it using taylor series:
$$\lim \limits_{x \to 0} \frac{\cos{x}\sin{x}-x}{\sin^3{x}}$$
$$\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}...$$
$$\cos{x}=1-\frac{x^2}{2!}+\frac{x^4}{4!}...$$
The problem I'm facing is how many terms to take when plugging in each taylor series into the original limit. I read in another thread that the lowest exponent in the denominator that I'll get is $x^3$, which means the largest degree I should have in the numerator is also $x^3$. Doing this I get:
$$\frac{\left(1-\frac{x^2}{2}\right)\left(x\right)-x}{x^3+...}= \frac{\frac{x^3}{2}}{x^3+...}$$
Of course I get the limit is $-\frac12$ which is incorrect. What did I do wrong?
Please, I'd prefer hints and a correction rather than a full blown answer.
 [/INST] You did not expand each element up to the same order in their Taylor series :
$$\cos (x)=1-\frac{x^2}{2!}+o(x^3)$$
$$\sin(x)=x-\frac{x^3}{3!}+o(x^3)$$
$$(\sin(x))^3=x^3+o(x^3)$$
Now we have : $$\lim \limits_{x \to 0} \frac{(\cos x)(\sin x)-x}{(\sin x)^3}=\lim \limits_{x \to 0} \frac{x-\frac{x^3}{3!}-\frac{x^3}{2!}-x+o(x^3)}{x^3+o(x^3)}=\lim \limits_{x \to 0} \frac{-\frac{2x^3}{3}+o(x^3)}{x^3+o(x^3)}=\lim \limits_{x \to 0} \frac{-2x^3}{3x^3}=-\frac{2}{3}$$
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