Taylor expansion at the origin

Question

I am confused about the following question: $\ln(1+x)=x+xg(x)$ for $x$ in a neighborhood of $0$, where $g$ is a continuous function such that $g(0)=0$. I think this may have some relation to Taylor expansion, since $\ln(1+x)=x+O(x^2)$ when I do the expansion at $0$. But I could not figure out why one can get this result $\ln(1+x)=x+xg(x)$.

Answer 1

Note the taylor expansion of $ln(1+x)$ is:

$$\sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^k}{k} = x-\frac{x^2}{2}+\frac{x^3}{3}-....\\ = x+ x\left(-\frac{x}{2}+\frac{x^2}{3}-....\right)\\ = x+x\underbrace{\left(\sum_{k=1}^{\infty}(-1)^{k}\frac{x^k}{k+1}\right)}_{g(x)}$$

Answer 2

This is called the Peano form of remainder: $$\log(1+x)=x+o(x)$$ Where $f(x)=o(x)$ means that $$\lim_{x \to 0} \frac{f(x)}{x}=0$$