Find $\lim_{x \to 0^{+}} \frac{\ln(x+1)}{x}$ without using L'Hopital's Rule? Without using L'Hopital's Rule, how does one find $\lim_{x \to 0^{+}} \frac{\ln(x+1)}{x}$?

I was hoping to find a way using basic calc I, pre-differentiation knowledge and not knowing the definition of $e$--much like you can prove $\lim_{x\to\infty} \frac{\sin x}{x} = 1$ using a geometric argument and the squeeze rule/theorem.
If you know that $\lim_{x\to\infty}\left(1+\frac1x\right)^x=e$, either because this is how $e$ was defined or because you’ve already proved it, you can begin by letting $L=\lim_{x\to 0^+}\frac{\ln(1+x)}x$. The exponential function is continuous, so 

$$e^L=e^{\lim_{x\to 0^+}\frac{\ln(1+x)}x}=\lim_{x\to 0^+}e^{\frac{\ln(1+x)}x}=\lim_{x\to 0^+}(1+x)^{1/x}=\lim_{x\to\infty}\left(1+\frac1x\right)^x=e\;,$$

and hence $L=\ln e=1$.
You could use the series expansion: $$\ln(x+1)=x-x^2/2+x^3/3-x^4/4\ .\ .\ .$$