Solving $\lim_{x\to 0}\frac{\ln(1+x)}x$ without De L'Hospital 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.
When $u \rightarrow \infty$, $x = \frac 1u \rightarrow 0$, as needed.