Solve this limit using equivalent infinitesimals I am given the following limit, which I'm asked to solve.
$$L=\lim_{x\to 0}\dfrac{2(\tan x-\sin x)-x^3}{x^5}$$
I get confused with equivalent infinitesimals most of the time. I know the correct answer to this limit is $\boxed{1/4}$. However, this is what I did:
$$\left.
\begin{array}{l}
&\sin x \approx x\\
&\tan x \approx x
\end{array}
\right\}
\Rightarrow L=-\lim_{x \to 0} \dfrac{x^3}{x^5}=-\lim_{x \to 0} \dfrac{1}{x^2}=\boxed{-\infty}$$
I know this problem can indeed be solved by taking 3 terms in the MacLaurin Series of each function (i.e. for both the sine and tangent functions). But sometimes this is not always necessary. Sometimes it's sufficient to take the first non-zero term in its MacLaurin Series.
But as far as I was taught in school, the equivalent infinitesimal is defined as the first non-zero term in the MacLaurin Series of a function.
How many terms should I grab to go safe for every case? Why doesn't it suffice to take just the 1st non-zero term?
 A: The denominator is $x^5$, so you need to go up to degree $5$ in the numerator. There is no hard and fast rule like “take three terms just in case”.
Think to
$$
\lim_{x\to0}\frac{x-\sin x}{x^3}
$$
You can't stop at degree $1$, because this would just give
$$
\lim_{x\to0}\frac{o(x)}{x^3}
$$
that doesn't carry sufficient information for computing the limit. If you go to degree $3$, you get
$$
\lim_{x\to0}\frac{x-(x-x^3/6+o(x^3)}{x^3}=
\lim_{x\to0}\left(\frac{1}{6}+\frac{o(x^3)}{x^3}\right)
$$
which instead carries the requested information, because $\lim_{x\to0}\frac{o(x^3)}{x^3}=0$.
If the denominator is $x^k$ you generally have to go to degree $k$ also in the numerator.
In your case:


*

*$\displaystyle\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^5)$

*$\displaystyle\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+o(x^5)$
and therefore
\begin{align}
2(\sin x-\tan x)-x^3
&=2x+\frac{2x^3}{3}+\frac{4x^5}{15}-2x+\frac{x^3}{3}-\frac{x^5}{60}-x^3+o(x^5)
\\[4px]
&=\frac{x^5}{4}+o(x^5)
\end{align}
Like before, your limit is $1/4$.
You can stop before degree $5$ if terms with lower degree don't cancel, so you can do it “incrementally”; in this case, terms with degree $1$ cancel as well as terms with degree $3$ (no even degree term).
A: $\tan(x)\sim x$ and $\sin x\sim x$ imply that
$$\lim_{x\to 0}\frac{\tan(x)-\sin x}{x}=0,$$
which means that the difference $\tan(x)-\sin x$ goes to zero faster than $x$. But what about if we compare $\tan(x)-\sin x$ with the faster infinitesimal $x^3$? At this stage we don't know. Here we need more information about the behaviour of $\tan(x)-\sin x$ as $x\to 0$. The exponent $5$ at the denominator says that we should use the same "precision" also at the numerator. So we need the expansion of sin(x) and tan(x) up to the order 5.
Consider the following similar example. We have that 
$$\lim_{x\to 0}\frac{(x+ax^5)-(x-x^3)-x^3}{x^5}=a$$
but we are not able to find the limit if we simply use $x+ax^5\sim x$ and $x-x^3\sim x$. 
