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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.

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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.

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You cannot replace $\sin x$ by $x$ in (2). It only holds in $\times $ or $\div$, i.e.,

$\lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{h(x)}{g(x)}$ if $f(x) \sim h(x)$ as $x\to x_0$; $\lim_{x\to x_0}f(x)g(x)=\lim_{x\to x_0} h(x)g(x)$ if $f(x) \sim h(x)$ as $x\to x_0$;

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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 $$f(x)=\frac{1}{3}+\frac{17 x^2}{90}+O\left(x^4\right)$$

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Thanks for a very good question which focuses on the conceptual issues faced by students when learning limits. Unless one is very experienced in the art of calculus and analysis it is best to avoid such replacement in solving limit problems. These replacements are not justified by any theorem in calculus. An expert can however know when such replacement will not create a problem and when it will lead to inconsistencies.

I will discuss a bit in detail the specific replacement of $\sin x $ by $x$. The intuitive logic behind this replacement is the following fact $$\lim_{x \to 0}\frac{\sin x}{x} = 1$$ However this does not mean that the expression $\sin x$ is almost equal to $x$ in all contexts when $x$ is near $0$. What it really means is that the expression $\left(\lim\limits_{x \to 0}\dfrac{\sin x}{x}\right)$ can be replaced by number $1$. One should not try to attach any more meaning to it. Now we can see how to use this fact in a smart way for the current problem. We can proceed as follows $$\begin{aligned}L &= \lim_{x \to 0}\frac{\tan x - x}{x^{2}\sin x}\\ &= \lim_{x \to 0}\dfrac{\tan x - x}{x^{3}\cdot\dfrac{\sin x}{x}}\\ &= \lim_{x \to 0}\dfrac{\tan x - x}{x^{3}}\cdot\dfrac{1}{\dfrac{\sin x}{x}}\\ &= \lim_{x \to 0}\dfrac{\tan x - x}{x^{3}}\cdot\lim_{x \to 0}\dfrac{1}{\dfrac{\sin x}{x}}\\ &= \lim_{x \to 0}\dfrac{\tan x - x}{x^{3}}\cdot\dfrac{1}{\lim\limits_{x \to 0}\dfrac{\sin x}{x}}\\ &= \lim_{x \to 0}\dfrac{\tan x - x}{x^{3}}\cdot\dfrac{1}{\lim\limits_{x \to 0}\dfrac{\sin x}{x}}\\ &= \lim_{x \to 0}\dfrac{\tan x - x}{x^{3}}\cdot\dfrac{1}{1}\text{ (note the replacement)}\\ &= \lim_{x \to 0}\dfrac{\tan x - x}{x^{3}}\end{aligned}$$ If we carefully see the first and last expression in the above derivation it will look as if the $\sin x$ in denominator has been replaced by $x$ but in reality we have only replaced $\left(\lim\limits_{x \to 0}\dfrac{\sin x}{x}\right)$ with $1$ which is valid because these two things are equal. An expert in calculus knows (via the above detailed steps) that he can replace $\sin x$ by $x$ in some contexts when $x \to 0$ and thus reduces the above long derivation with a simple replacement.

But we must bear in mind that any such replacement must be justified by a long derivation as above which makes use of limit theorems and some standard limit formulas. By the way the advantage of a such a replacement is quite obvious. In the above problem if we apply L'Hospital then we can see that it is much easier to differentiate $x^{3}$ in denominator compared to the $x^{2}\sin x$.

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