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: 
By the definition of the derivative at $x=0$
$$
\log(1+x)=x+o(x)
$$
Hence
$$
\log(1+x)/x\to1.
$$
as $x\to0$.
$\dfrac{\log(1+x) -\log(1)}{x} =$

$ \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.