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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?

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    $\begingroup$ Use a backslash "\" when in doubt, in MathJax - placing backslashes before a standard expression such as $\lim$ will usually be rendered properly: e.g., $\lim$, $\sin$ render as $\lim$ and $\sin$ instead of $lim$ and $sin$ $\endgroup$
    – FShrike
    Commented Sep 19, 2021 at 11:10
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    $\begingroup$ Since the limit has one distinct answer, it would be very useful for your question if you showed how you obtained $\frac{1}{3}$, and how using a different little-o would supposedly lead to a different answer $\endgroup$
    – FShrike
    Commented Sep 19, 2021 at 11:12
  • $\begingroup$ @FShrike if I use the first series I worte, I get that the function within the brakets in the limit becomes $\frac{x^2+\frac{x^3}{3}+o(x^4)}{x^3}-1/x$ and when I do the division, I get $1/x+1/3+ o(x^4)/x^3-1/x$ which goes to 1/3 as x approaches 0. If I stop at $o(x^3)$, I don't get the 1/3 term because I simply have $o(x^3)/x^3$ which is 0 $\endgroup$ Commented Sep 19, 2021 at 11:18

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$$\lim_{x\to0}2\frac{e^x-1-x}{x^3}-\frac{1}{x}=\lim_{x\to0}\frac{1}{x}+\frac{1}{3}+\frac{1}{12}x+\cdots-\frac{1}{x}=\frac{1}{3}$$

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.

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    $\begingroup$ Now I understand, thanks! $\endgroup$ Commented Sep 19, 2021 at 11:33

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