How can we determine the number of terms which we have to take in a series to get a particular accurate? As I remember , two days ago , there was a question ( here ) asks for calculating this limit 
$\displaystyle \lim \limits_{x\rightarrow \infty } \frac{x^3}{e^x}$ and the question was answered . 
of course this is an easy limit , we can calculate it using l'hopital rule . 
Now ,  if I expressed $\displaystyle e^x = 1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots  = \sum\limits_{i=0}^{\infty}\frac{x^i}{i!}$ and substitute back an approximation for $e^x$. 
If we approximated $e^x$ for 3 terms then the limit is $\infty$ , if we approximated it for 4 terms then the limit is 6 but if we approximated for 5 terms then we get the right answer which is 0 . 
Now , How can we know that approximation for 5 terms will work while approximation for 4 or 3 terms would fail ? Is there any test or analytical method to determine this ? 
I used the geometric approach to see the reason behind that . here is the sketch for those approximations and the original function which illustrate and show why approximation for  3 and 4 terms fail . but I ask for Analytical approach as using the sketch of approximations is not rigorous and practical in general . 

 A: The problem is that the sum doesn't really approximate $e^x$ "in a neighborhood of $\infty$". The nicer argument is that for any $n$ that we pick $$e^x=\sum_{k\geqslant 0}\frac{x^k}{k!}\geqslant \frac{x^n}{n!}$$
Thus for any monomial $x^m$ we can simply pick $n>m$ and write $$\frac{x^m}{e^x}\leqslant n!{x^{m-n}}\to 0$$
as $x\to\infty$. The argument of omitting terms is just ill founded if you're looking for a "tight" approximatio $-$ else, you're gust getting crude overestimates. You can "omit" some terms when $x\to 0$, but here we're looking at $x\to\infty$.
A: It took me a while to figure out what the question was asking. It seems the question is why
$$
\begin{align}
\lim_{x\to\infty}\frac{x^3}{1+x+\frac{x^2}{2}}&=\infty\tag{1}\\
\lim_{x\to\infty}\frac{x^3}{1+x+\frac{x^2}{2}+\frac{x^3}{6}}&=6\tag{2}\\
\lim_{x\to\infty}\frac{x^3}{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}}&=0\tag{3}\\
\end{align}
$$
while
$$
\lim_{x\to\infty}\frac{x^3}{e^x}=0\tag{4}
$$
In actuality, $e^x$ grows faster than any power of $x$. We see why in the example of $x^3$ above. First of all, for $x\ge0$
$$
e^x\ge1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots+\frac{x^n}{n!}\tag{5}
$$
for any $n$.

Thus, $(5)$ (substituting $n\mapsto n+1$) implies
$$
\begin{align}
\lim_{x\to\infty}\frac{x^n}{e^x}
\le&\lim_{x\to\infty}\frac{x^n}{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots+\frac{x^n}{n!}+\frac{x^{n+1}}{(n+1)!}}\\
=&\lim_{x\to\infty}\frac1{\frac1{x^n}+\frac1{x^{n-1}}+\frac1{2x^{n-2}}+\frac1{6x^{n-3}}+\dots+\frac1{n!}+\frac{x}{(n+1)!}}\\
\le&\lim_{x\to\infty}\frac{(n+1)!}{x}\\[14pt]
=&0\tag{6}
\end{align}
$$

If we'd used one fewer term in the denominator of $(6)$, we would get
$$
\begin{align}
\lim_{x\to\infty}\frac{x^n}{e^x}
\le&\lim_{x\to\infty}\frac{x^n}{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots+\frac{x^{n-1}}{(n-1)!}+\frac{x^n}{n!}}\\
=&\lim_{x\to\infty}\frac1{\frac1{x^n}+\frac1{x^{n-1}}+\frac1{2x^{n-2}}+\frac1{6x^{n-3}}+\dots+\frac1{x(n-1)!}+\frac1{n!}}\\[4pt]
=&n!\tag{7}
\end{align}
$$
because we are using a polynomial of the same order as the numerator to approximate $e^x$ near $x=\infty$.

If we'd used two fewer terms in the denominator of $(6)$, we would get
$$
\begin{align}
\lim_{x\to\infty}\frac{x^n}{e^x}
\le&\lim_{x\to\infty}\frac{x^n}{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots+\frac{x^{n-2}}{(n-2)!}+\frac{x^{n-1}}{(n-1)!}}\\
=&\lim_{x\to\infty}\frac1{\frac1{x^n}+\frac1{x^{n-1}}+\frac1{2x^{n-2}}+\frac1{6x^{n-3}}+\dots+\frac1{x^2(n-2)!}+\frac1{x(n-1)!}}\\
=&\lim_{x\to\infty}x\cdot\lim_{x\to\infty}\frac1{\frac1{x^{n-1}}+\frac1{x^{n-2}}+\frac1{2x^{n-3}}+\frac1{6x^{n-4}}+\dots+\frac1{x(n-2)!}+\frac1{(n-1)!}}\\
=&\lim_{x\to\infty}x\cdot(n-1)!\\[9pt]
=&\infty\tag{8}
\end{align}
$$
because we are using a polynomial of a lower order than the numerator to approximate $e^x$ near $x=\infty$.

However, given $(6)$, $(7)$, and $(8)$, we can easily conclude that
$$
\lim_{x\to\infty}\frac{x^n}{e^x}\le\min(0,n!,\infty)=0\tag{9}
$$
