<s>[INST] Questions about the computation of a limit. By L'Hôpital's rule, it is easy to see that 
$$
\lim_{x \to 0} \frac{\tan x - x}{x^2 \sin x} = 1/3.
$$
But if we use the following procedure to compute the limit, we get a wrong answer. The procedure is in the following. 
$$
\lim_{x \to 0} \frac{\tan x - x}{x^2 \sin x} = 
\lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - x}{x^2 \sin x}. \quad (2)
$$
Since $\sin x \sim x$ ($\sin x$ and $x$ are equivalent infinitesimals), we replace $\sin x$ by $x$ in (2). Then we obtain
$$
\lim_{x \to 0} \frac{\tan x - x}{x^2 \sin x} = 
\lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - x}{x^2 \sin x} \\
= \lim_{x \to 0} \frac{\frac{x}{\cos x} - x}{x^3} \\
= \lim_{x \to 0} \frac{\frac{1}{\cos x} - 1}{x^2} \\
= \lim_{x \to 0} \frac{1 - \cos x}{x^2 \cos x} \\
= 1/2.
$$
I don't konw where is the problem. Thank you very much.
 [/INST] Another solution could be to double check the result using Taylor series $$f(x)=\frac{\tan x - x}{x^2 \sin x}$$ and use sucessively $$\tan(x)=x+\frac{x^3}{3}+O\left(x^4\right)$$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ So, the numerator is $$\tan(x)-x=\frac{x^3}{3}+O\left(x^4\right)$$ and denominetaor is $$x^2\sin(x)=x^3-\frac{x^5}{6}+O\left(x^6\right)$$ and then the limit of $\frac{1}{3}$.
Personally, what I enjoy with Taylor series is that they show how the limit is reached. In this case, close to $x=0$, we obtain $$f(x)=\frac{1}{3}+\frac{17 x^2}{90}+O\left(x^4\right)$$
</s>