calculus - 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.

Answer

What you did:
$${\sin x} \sim x \implies \frac{\sin x}{\cos x} - x \sim \frac{x}{\cos x} - x$$
is wrong, because you added equivalents: you can multiply by $1/\cos x$, to get
$${\sin x} \sim x \implies \frac{\sin x}{\cos x} \sim \frac x{\cos x}$$

but you can't fo the manipulation with the sum:
$$\frac{\sin x}{\cos x} \sim \frac x{\cos x} \nRightarrow \frac{\sin x}{\cos x} - x \sim \frac{x}{\cos x} - x$$

---

Even when $f\sim g$, you can be in the case for which
$$g+h \nsim f+h$$

For example, $f(n) = n^2 - n$, $g(n) = n^2$, $h(n) = 1-n^2$.

This is true in particular when:

• $f\sim a y$, $h\sim b y$ for a function $h$ and scalars $a,b$ with $a+b \neq 0$


• $h = o(g)$.

Otherwise, do not consider making the sum (or substitutions, which is the same) of equivalents.