Choosing the "right order" of approximation in Taylor Series I was evaluating the following limit using Taylor series, $\lim_{x \rightarrow 0}(2 \frac{e^x-1-x}{x^3}-1/x)$, and I wrote the exponential function as $e^x=1+x+\frac{x^2}{2}+ \frac{x^3}{6}+o(x^4)$ and got that the limit is $\frac13$. But, if I had written $e^x=1+x+\frac{x^2}{2}+o(x^3)$, I would have obtained a different result, so how do I choose correctly at which order I'm supposed to stop when writing a Taylor series to evaluate a limit?
 A: $$\lim_{x\to0}2\frac{e^x-1-x}{x^3}-\frac{1}{x}=\lim_{x\to0}\frac{1}{x}+\frac{1}{3}+\frac{1}{12}x+\cdots-\frac{1}{x}=\frac{1}{3}$$
However, if you choose to use little-o notation, recall that the definition of little-0 is that:
$$f(x)\in\omicron(g(x)),\,x\to0\iff\lim_{x\to0}\frac{f(x)}{g(x)}=0$$
Whereas the definition of big-O notation is:
$$f(x)\in O(g(x)),\,x\to0\iff\lim_{x\to0}\frac{f(x)}{g(x)}\lt\infty$$
So we notice that:
$$\frac{x^3}{6}+\frac{x^4}{24}+\cdots\in O(x^3)$$
But is not little-o of $x^3$ - this is because the remainder terms divided by $x^3$ leave you with $\frac{1}{6}$ and some other terms that go to zero, leaving $\frac{1}{6}\neq0$ as the limit.
So in fact $e^x=1+x+\frac{x^2}{2}+O(x^3)$, not little-0!
Now:
$$\begin{align}\lim_{x\to0}2\frac{e^x-1-x}{x^3}-\frac{1}{x}&=\lim_{x\to0}2\frac{\frac{x^2}{2}+O(x^3)}{x^3}-\frac{1}{x}\\&=\lim_{x\to0}2\cdot\frac{O(x^3)}{x^3}\end{align}$$
Which is an ambiguous expression that is not necessarily (and not actually, in this case) equal to $0$. If we replaced $O(x^3)$ with: $\frac{1}{6}x^3+o(x^3)$ however, that would be correct, as the remainder terms when divided by $x^3$ are $\frac{1}{24}x+\frac{1}{120}x^2+\cdots$ which do go to zero as $x$ does.
To demonstrate the proper use of these notations, this is what you should be doing:
$$\lim_{x\to0}2\frac{e^x-1-x}{x^3}-\frac{1}{x}=\cdots=\lim_{x\to0}\frac{1}{3}+\frac{o(x^3)}{x^3}=\frac{1}{3}$$
Since $o(x^3)/x^3\to0$ by definition.
