I have 2 questions about using Taylor series to solve limits,
(1) I have $\log(x^5)$, I think I cannot use Taylor series to approximate this $\log$, because $\log$ doesn't tend to $1$ as $x$ goes to $0$. I think this problem could be solved this way: just adding and subtracting $1$, as follows:
$$\log(1+ (x^5-1) )$$
but now I have another problem, because $(x^5-1)$ doesn't tend to $0$.
(2) I have $\log(2+e^x)$; it's like the previous $\log$, because I can't use the trick of adding $1$ and subtracting $1$. Therefore, I've tried to collect like terms even if I don't have common terms, as follows:
$$\log(e^x(1 + 2/e^x))$$
but that's still not correct, because when I plug zero in the function, I got that $e^0 = 1$, so I have $1 + 2 = 3$; but it's not what I want. the function has to look like this: $\log(1+ \text{something})$, and $\text{something}$ must tend to $0$, as $x$ approaches $0$.
Assuming that $x$ tends to zero,
You can write $\log(x^5) = 5 \log(x)$. This cannot be approximated by polynomials or rational functions of $x$, but such approximation is not needed. One uses results such as $x^\alpha |\log(x)|^\beta \to 0$, for $\alpha > 0$.
One can write $\log(2 + e^x) = \log(3) + \log(1 + \frac{e^x-1}{3})$, which has the form $\log(1 + u)$ with $u\to 0$.
In general, if you have $\log(\varphi(x))$ and $\varphi(x)\to \ell > 0$, you can write \begin{equation} \log(\varphi(x)) = \log(\ell) + \log\left(1 + \frac{\varphi(x)-\ell}{\ell}\right) \end{equation} which has the form $\log(1 + u)$ with $u\to 0$.