Question: I want to prove that $\lim\_{x\to 0}\frac{\ln (1+x)}{x}=1$ by Taylor series. I have get to the step: $$\lim\_{x\to 0}\frac{\ln (1+x)}{x}=\lim\_{x\to 0}\frac{x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots}{x}$$

Next, when $x\neq0$, $\frac{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots}{x}=\frac{1}{x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\right)=1-\frac{x}{2}+\frac{x^2}{3}-\cdots$. This is valid because $\sum\_{n=1}^\infty(-1)^{n+1}\frac{x^n}{n}$ converges for every $x\neq 0$ near $0$, and then we can multiply a constant $\frac{1}{x}$ inside it. However, how can I rigorously and easily say that $\lim\_{x\to 0}(1-\frac{x}{2}+\frac{x^2}{3}-\cdots)=1$?

Answer: 
You can be pretty brutal about this and just expand with the triangle inequality:

\begin{align\*}
\left\lvert x - \frac{x^2}2 + \frac{x^3}3 - \cdots\right\rvert \le x + x^2 + x^3 + \cdots = \frac{x}{1 - x} \to 0
\end{align\*}

as $x \to 0$.

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That being said, it sorta misses the point to use Taylor series like this\*. The limit you're looking at is the definition of
$$\left.\frac{d}{dx}\bigl(\ln(1 + x)\bigr)\right\vert\_{x=0}$$
which you would need to compute in order to even know the Taylor series.

\* If you've only defined log via Taylor series, or every used the derivative of log to compute the Taylor series.
$\dfrac{\log(1+x) -\log(1)}{x} =$

$ \log'(t) =1/t,$ $1< t<1+x.$

$\lim\_{x \rightarrow 0} (1/t) =1.$

$\lim\_{x \rightarrow 0} \dfrac{\log(1+x)-\log(x)}{x} = 1.$

Used: Mean Value Theorem.