$\ln\left(1+x\right)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dots$ when $|x|<1$ or $x=1$. $\ln\left(1+x\right)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dots$ when $|x|<1$ or $x=1$. Why is the restriction $|x|<1$ or $x=1$? I know from Wikipedia that it is because out of this restriction, the function's "Taylor approximation" is not fair. But how did we derive this limit?

I think your question is why $|x|<1$ and x=1. The reason is that the formula is not valid for $x=-1$. So we can't say $|x| \le 1$ as we want to exclude $x = -1$.
The limit has to be less than $1$ as $\log(1+x)$ blows up at $-1$. To show that we can get this close to $1$ requires more analysis.
