<s>[INST] Why using fewer terms of Taylor series doesn't give $0/0$ but gives a wrong answer? I was reading a Calculus book and I saw this problem which looks easy:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} = ?$$
It's a 0/0 limit and it's using some of the Taylor series of $\sin$ and $\cos$ expressions to solve the problem.
I know that the First and Second way should be correct because it's using more expressions of the Taylor series around 0.
What I can't figure out is WHY using fewer expressions of the Taylor series in the Third way doesn't give 0/0 but gives a wrong answer?
First way:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x-2 \sin x \cos x}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x(x-\sin x)}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x\left(x-x+\frac{x^3}{6}\right)}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x\left(\frac{x^3}{6}\right)}{x^3}=\frac{1}{3}$$
Second way:
$$\lim _{x \rightarrow 0} \frac{2x(1-\frac{x^2}{2})-(2x-\frac{8x^3}{6})}{x^3}=\lim _{x \rightarrow 0} \frac{2x-x^3-2x+\frac{8x^3}{6}}{x^3}=\lim _{x \rightarrow 0} \frac{\frac{x^3}{3}}{x^3}=\frac{1}{3}$$
Third way:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} =\lim _{x \rightarrow 0} \frac{2 x \cos x-2x}{x^3}=\lim _{x \rightarrow 0} \frac{2x(\cos x -1)}{x^3}=\lim _{x \rightarrow 0} \frac{2x(-\frac{x^2}{2})}{x^3}=-1$$
 [/INST] In all the cases we should use remainder to proceed properly as follows
$$\frac{2 x \cos x- \sin 2x}{x^3}=\frac{2 x \left(1-\frac12 x^2+O(x^3)\right)-  \left(2x-\frac16 (2x)^3+O(x^4)\right)}{x^3}=$$
$$=\frac{2x-x^3-2x+\frac43x^3+O(x^4)}{x^3}=\frac13+O(x) \to \frac13$$
without remainder we can easily get wrong with the solution.
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