Trying to solve this limit without derivatives I found this answer that is pretty straightforward and I can easily follow the flow. I can understand why ${u\to \infty}$ because:
$$\lim_{u\to\infty}(1 + \frac{1}u)^u = e $$
but how there is a relation to ${x\to 0}$ with ${u\to \infty}$ when we need to find the limit that approaches $0$.
We can squeeze $\ln(1+x)$ between integrals.
For upper bound we use $\frac 1{1+t}\le \frac 1{1+t_{min}}=1$
For lower bound $\frac 1t\ge \frac 1{t_{max}}=\frac 1{1+x}$.
$$\dfrac x{1+x}=\int_1^{1+x}\dfrac{\mathop{dt}}{1+x}\le\int_1^{1+x}\dfrac{\mathop{dt}}{t}=\ln(1+x)=\int_0^x\dfrac{\mathop{dt}}{1+t}\le\int_0^x\mathop{dt}=x$$
Now dividing by $x$ we get
$$\dfrac 1{1+x}\le\dfrac{\ln(1+x)}x\le 1$$
Since both sides converge to $1$, this is also the limit for our ratio.
$$A=\lim_{x\to 0}\frac{\ln(1+x)}x$$ Substitute $u=\frac1x$, so the limit becomes $u\to \infty$. $$A=\lim_{u\to\infty}\left\{u\ln\left(1+\frac1u\right)\right\}=\lim_{u\to\infty}\left\{\ln\left(1+\frac1u\right)^u\right\}=\ln\left(\lim_{u\to\infty}\left\{\left(1+\frac1u\right)^u\right\}\right)=\ln(e)=1$$
$$\lim_{x\to 0}\frac{\ln(1+x)}x=\lim_{x\to 0} \frac{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots}{x}=1$$
When $u \rightarrow \infty$, $x = \frac 1u \rightarrow 0$, as needed.
