For the series $e^{\tan x}$, estimate the series finding the best integer value possible using o- and O-notation. Consider the series:
$$e^{\tan x} = 1+x+\frac{x^2}{2!}+\frac{3x^3}{3!}+\frac{9x^4}{4!}+\quad...\qquad  (|x|\le\frac{\pi}{2})$$
Retaining three terms, estimate the remaining series with o-notation for the best integer value possible, as $  x \to 0 $.
(Source: Numerical Analysis: Mathematics of Scientific Computing, Chapter 1.2)
I know that the answer should look like $$e^{\tan x} = 1+x+\frac{x^2}{2!}+o(x^k)$$ such that $$e^{\tan x} -( 1+x+\frac{x^2}{2!})=o(x^k)$$ 
with $k$ being the largest integer n that makes the statement true, since $  x \to 0 $. In my course, we learned that a function is little o if $\left|\frac{f(x)}{x^k}\right|\le0$ as $  x \to 0 $.
My question is exactly how to determine the "best" value k to estimate this function? Also, I am aware that this question is already posted but I did not find the answers to be sufficient. Any insight would be greatly appreciated!!
 A: The definition of your little o, does not make sense, as Robert Israel mentioned in his comment. Here are the definitions.
A function $g(x)=o(h(x)),\quad (x→a)$ if: $$\lim_{x→a}\frac{|g(x)|}{|h(x)|}=0$$
A function $g(x)=\mathcal{O}(h(x)),\quad (x→a)$ if: $$\lim_{x→a}\frac{|g(x)|}{|h(x)|}=c<∞¹$$
As you already stated, we need to subtract the first three terms.
$$F:=e^{\tan x} - ( 1+x+\frac{x^2}{2!}) =\frac{3x^3}{3!}+\frac{9x^4}{4!}+\quad...$$
So we get: 
$$1)\qquad\frac{F}{x^1}= \frac{3x^2}{3!}+\frac{9x^3}{4!}+…\quad  → 0\quad (x→0) $$
$$2)\qquad\frac{F}{x^2}= \frac{3x}{3!}+\frac{9x^2}{4!}+…\quad  → 0\quad (x→0) $$
$$3)\qquad\frac{F}{x^3}= \frac{3}{3!}+\frac{9x}{4!}+…\quad  → \frac{3}{3!}\quad (x→0) $$
$$4)\qquad\frac{F}{x^4}= \frac{3}{3!x}+\frac{9}{4!}+…\quad  → ∞\quad (x→0) $$
Hence: $$F=\mathcal{O(x^3)}, \qquad F=o(x^2)$$
For small-$o$ you have two options: $o(x)$ and $o(x^2)$.  Here "best" means largest $k$ of $o(x^k)$, since it gives you more information.
It gives you more information, as $x^m=o(x^k)$, $(x→0)$ for $m>k$ $(\star)$.
So $x^2=o(x)$, but $x\neq o(x^2)$.
Hence: \begin{align*}F=o(x^2) &⇒ F=o(x) \\ F=o(x) &\not⇒ F=o(x^2)\end{align*}
Therefore $F=o(x^2)$ 'holds more information'.
(That is also the reason, why I didn't compute $F/x^k$, $k=0,-1,…$.)
Note that "best" does not always mean largest in the context of Landau-Symbols, for example if you look at $x→∞$ instead of $x→0$, you would want the smallest.²
Now for $\mathcal{O}$ you get $F=\mathcal{O}(x^3)$ using the same argument. When speaking about Taylor-series, it usually is $k_{\mathcal{O}}=k_o+1$, since you express a sum of monomials as $\mathcal{O}/o$ of another monomial.

¹ To be exact it would be $\lim\sup$ there. But I have never seen an example, where you really need that.
² In most cases you are interested in either $x→0$ or $x→∞$. I can't think of an interesting case $x→a$ at the moment.
$h→0$ is interesting for example in the context of convergence of a solution $u_h$ to $u$ of a partial differential equation. There $h$ represents the size of the grid. Another example, where you also look at Taylor-series, are the convergence (consistency) theorems of numerical methods for ordinary differential equations. There $h$ is the step-size.
$n→∞$ is interesting when it comes to the computation time of an algorithm seen as function of the size ($n$) of the input data, e.g. solving linear systems in $\mathcal{O}(n^3)$, with $n$ being the size of the system matrix.
$(\star)$ Exercise: Proof that statement!
