$\lim_{x\to0} \frac{\ln(x+e)-1}{x}$ I want to calculate the limit $\lim_{x\to0} \frac{\ln(x+e)-1}{x}$.
Just putting in the $0$ we can see $\frac{\ln(e)-1}{0} = \frac{0}{0}$ so my first guess is to use "L'Hopital" rules.
$\lim_{x\to0} \frac{\ln(x+e)-1}{x} = \lim_{x\to0} \frac{\frac{1}{x+e}}{1} = \frac{1}{e}$.
edit: I think this is the correct solution, can anyone approve?

This works. If you're interested, an alternative way to attack this problem is to recognize that the limit
$$\lim_{x\to 0}\frac{\ln(x+e)-1}{x}=\lim_{x\to 0}\frac{\ln(x+e)-\ln(e)}{x}$$
is the derivative of $\ln$ evaluated at $e$. You know that $\frac{d}{dx}\ln(x)=\frac{1}{x}$ for every positive $x$, so
$$\lim_{x\to 0}\frac{\ln(x+e)-1}{x}=\frac{1}{e}$$
