TITLE: How can I deduce that $\lim\limits_{x\to0} \frac{\ln(1+x)}x=1$ without Taylor series or L'Hospital's rule?
QUESTION [5 upvotes]: How can I calculate this limit without Taylor series and L'Hospital's rule?
$$\lim          _{x\to0} \frac{\ln(1+x)}x=1$$

REPLY [7 votes]: If the limit exists, say $L$, then you can state that:
$$\begin{align}
L&=\lim_{x\to0}\frac{\ln(1+x)}{x}\\
\therefore e^L&=e^{\lim_{x\to0}\frac{\ln(1+x)}{x}}\\
&=\lim_{x\to0}e^{\frac{\ln(1+x)}{x}}\\
&=\lim_{x\to0}(e^{\ln(1+x)})^\frac{1}{x}\\
&=\lim_{x\to0}(1+x)^\frac{1}{x}\\
&=e\\
\therefore L&=1
\end{align}$$