<s>[INST] Is this way of evaluating $\lim_{x\to 0}\frac{x-\sin x}{x^3}$ rigorous? Assume that you want to evaluate $$\lim_{x \to 0} \frac{x-\sin x}{x^3}.$$ Of course, you may apply $\textbf{de l'Hôpital's Rule}$, that's natural and simple. But what if you are required to avoid using the more advanced knowledge such as “de l'Hôpital”，“Taylor” and so on？Now,someone gives a solution for the problem as follows. 
Since$$\begin{align*}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}&=\lim\limits_{x \to 0} \dfrac{2x-\sin 2x}{(2x)^3}\\[4pt]&=\lim\limits_{x \to 0} \dfrac{2x-2\sin x\cos x}{(2x)^3}\\[4pt]&=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x+\sin x-\sin x\cos x}{x^3}\\[4pt] &=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}+\frac{1}{4}\lim\limits_{x \to 0} \dfrac{\sin x(1-\cos x)}{x^3}\\[4pt] &=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}+\frac{1}{4}\lim\limits_{x \to 0} \dfrac{\sin x}{x}\cdot\lim_{x \to 0}\frac{1-\cos x}{x^2}\\[4pt] &=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}+\frac{1}{4}\cdot 1\cdot \frac{1}{2}\\[4pt] &=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}+\frac{1}{8},\end{align*}$$ which sets up an equation on the limit, hence, by solving the equation, we obtain that$$\lim_{x \to 0} \frac{x-\sin x}{x^3}=\frac{1}{6}.$$
What about the solution? Is it rigorous? Hope to read your comments. Thank you!
 [/INST] There's a typo: when you wrote (after the third equality) $\sin\cos x$, you meant $\sin x\cos x$.
This proves correctly that if the limit exists then it must be equal to $\frac16$, but that's all.
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