Proof that $\lim_n\frac{\ln(1+x_n)}{x_n}=1$

Given a sequence of real numbers $x_n$ that converges to $0$ and $\vert x_n \vert <1$, $x_n \neq0$, I must prove the following limit:

$\lim_n\frac{\ln(1+x_n)}{x_n}=1$

What I have tried is:

$\lim_n\frac{\ln(1+x_n)}{x_n}=\lim_n\frac{1+x_n}{e^{x_n}}=\frac{1}{1}=1 $

However, I don't feel like this is right. How can I improve this?

Observe that $$\lim_{n\to\infty}\frac{\ln(1+x_n)}{x_n}=\lim_{n\to\infty}\frac{\ln(1+x_n)-\ln(0)}{x_n} =\lim_{h\to0}\frac{\ln(1+h)-\ln(0)}{h} $$ is the derivative of $\ln$ at $x=1$.

You can use L'Hôpital's rule, as the functions are differentiable and the derivative of the denominator is not zero: $$\lim_{n\to\infty} \frac{\text{log}(1+x_n)}{x_n} = \lim_{x\to 0} \frac{\text{log}(1+x)}{x} = \lim_{x\to 0} \frac{\frac{1}{1+x}}{1} = 1 .$$ The first equality follows simply as the functions are continuous, so in the limit they do not care about the specific sequence.