I have some problems understanding the following statement:
If we want that the sequence $(a_n)$ satisfies $(1-a_n)^n \rightarrow \text{e}^{-x}$, we have to find $a_n$ such that $a_n = x/n + o(1/n)$. Especially we can take $\tilde a_n = x/n$.
I know that $a_n = x/n + o(1/n)$ means $\lim_{n \to \infty} \frac{a_n-x/n}{1/n} = \lim_{n \to \infty} (n a_n - x) =0$. But why does this imply $(1-a_n)^n \rightarrow \text{e}^{-x}$?
Second, I do not understand why $\tilde a_n = x/n$ is a sequence with $\tilde a_n = x/n + o(1/n)$. This would imply $o(1/n)=0$, which is nonsense.
For the first point, this is because $$\log\bigl((1-a_n)^n\bigr)=n\log\biggl(1-\frac xn+o\Bigl(\frac1n\Bigr)\biggr)=n\biggl(-\frac xn+o\Bigl(\frac1n\Bigr)\biggr)=-x+o(1).$$ For the second point, you're misusing the symbol $o$. We can write $x/n=x/n+o(1/n)$ because $o(1/n)$ doesn't denote a specific function, but a class of functions. Actually, it would be more rigourous to write $\;x/n\in x/n+o(1/n)$.