Can we say that $x=0$ is a double root of $f(x) = (e^x-1)(\ln( x+1))$?

Question

Let $$f(x) = (e^x-1)(\ln( x+1))$$

So far , I've only seen examples of textbooks referencing repeated roots if we can write the function with linear factors raised to some natural exponent ( like if $a$ is a repeated root , then we can write the function as $(x-a)^{m+1}g(x), m\in\mathbb N$). But in above example, the factors are not linear polynomials.

So , Is it correct to say that $x=0$ is a repeated root of $f$ or not?

Context:

I'm confirming the terminology because currently I'm studying derivatives. My teacher said that functions having repeated roots are always differentiable at that point with derivative equal to $0$ . That's why I asked question whether above type of functions come into that category so that the property of differentiability could be extended.

Answer 1

Try to take the limit of $f(x)$ divided by $x^2$ at $x=0$. If it exists and it is different from 0, then $f(x)$ has a double root at $x=0$.

Answer 2

You can indeed say that $0$ is a repeated root of this function. More generally, if $f(x)$ and $g(x)$ are differentiable functions which both have a root at $x = a$, then $h(x) = f(x) g(x)$ has a repeated root there, because $h(a) = 0$ and $h'(a) = f(a) g'(a) + f'(a) g(a) = 0$.

Answer 3

$\displaystyle \lim_{ x\to 0}\frac{f(x)}{x^2}=1$ so $\text{ord}_{x=0}(f)=2 $, i.e., $f$ has a double zero in $x = 0$.

EDIT.
$$e^x-1= x\sum_{n=1}^{+\infty}\frac{x^{n-1}}{n!}$$ and
$$\ln(x+1)= x\sum_{n=1}^{+\infty}(-1)^{n-1}\frac{x^{n-1}}{n}$$
therefore $f(x)=x^2g(x)$ with $g(0) \neq 0$