Asymptotic equivalence with the Notable Limit of Natural Logarithm  I hope this is not too easy of a question
Basically I've been fiddling with Asymptotic Equivalences to solve limits in a faster and less prone to error way, and I am facing the limit:
$$
\lim_{x\to0}{\frac{e^x-\ln(e+x)}{x^3-2x}}
$$
The approach I wanted to use is that of substitution via Asymptotic Equivalence, aiming at substituting $e^x-1$ with $x$ and $\ln(e+x)$.
My idea was to add and subtract $1$ in the numerator in order to obtain the right situation to substitute $e^x-1$ with $x$; this decision probably isn't the best one to solve this limit in a definite manner since I'll remain with $\frac{1}{x^3-2x}$ that is of indefinite form but as I was trying out this path I started asking myself if it was possible to substitute $\ln{e+x}$ via Asymptotic Equivalences.
I was thinking that maybe
$$
\ln{\frac{1+f(x)}{f(x)}}=1\implies \ln{\frac{e+x}{x}}=e;
$$
am I completely wrong? If not can one of you point me to a resource (be it online or a book) in which I can delve deeper into Notable Limits and Asymptotic Equivalences?
 A: Clearly, when $x\to 0$, we have $x^3=o(x)$, so we can consider only $2x$.
The definition of $o(...)$ or "little-o": let $f(x),\, g(x)$ two real functions. Let $x_0 \in \overline{\mathbb{R}}$ an accumulation point for $g(x)$ and such that $g(x) \neq 0\,\, \forall x \,\in \,I(x_0)$. We write $f(x)=o(g(x))$ when $x\to x_0$ if and only if:
$$\lim_{x\to x_0}\frac{f(x)}{g(x)}=0$$
In this case, let $f(x)=x^3$, $g(x)=2x$ and $x_0=0$. We have:
$$\lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{x^3}{2x}=\lim_{x\to x_0}\frac{1}{2}x^2=0$$
Or, when $x\to 0$, we have $2x>x^3$. Here, we have the graph (in red $y=2x$ and in blue $y=x^3$):

Also, we know from asymptotic relations (when $x\to 0$):
$$e^x-1\,\,\sim\,\,x$$
And:
$$\log(1+x)\,\,\sim\,\,x$$
Now, we can approach your limit:
$$\lim_{x\to0}{\frac{e^x-\ln(e+x)}{x^3-2x}}\,\,\sim\,\,\lim_{x\to0}\frac{e^x-1-\ln\left(1+\frac{x}{e}\right)}{-2x}\,\,\lim_{x\to 0}\frac{x-\frac{x}{e}}{-2x}=-\frac{1-\frac{1}{e}}{2}=\frac{1-e}{2e}$$
A: With a bit of algebraic manipulation
$$
\frac{e^x-\log(e+x)}{x^3-2x}=\frac{\log(e+x)-e^x}{2x}\frac{1}{1-(x^2/2)}
$$
so that
$$
\lim_{x\to0}\frac{e^x-\log(e+x)}{x^3-2x}=\left(\lim_{x\to0}\frac{\log(e+x)-e^x}{2x}\right)\left(\lim_{x\to0}(1-x^2/2)^{-1}\right)
=\lim_{x\to0}\frac{1}{2}\frac{\log(e+x)-e^x}{x}.
$$
Then note that $e^x=1+x+\mathcal O(x^2)$ and $\log(e+x)=1+x/e+\mathcal O(x^2)$ as $x\to 0$; hence,
$$
\frac{\log(e+x)-e^x}{x}=\frac{1}{e}-1+\mathcal O(x)
$$
and
$$
\lim_{x\to0}\frac{1}{2}\frac{\log(e+x)-e^x}{x}=\lim_{x\to0}\frac{1}{2}\left(\frac{1}{e}-1+\mathcal O(x)\right)=\frac{1}{2}\left(\frac{1}{e}-1\right).
$$
