TITLE: Find the limit of $\lim_{x\to0}{\frac{\ln(1+e^x)-\ln2}{x}}$ without L'Hospital's rule
QUESTION [8 upvotes]: I have to find: $$\lim_{x\to0}{\frac{\ln(1+e^x)-\ln2}{x}}$$
and I want to calculate it without using L'Hospital's rule. With L'Hospital's I know that it gives $1/2$.
Any ideas?

REPLY [2 votes]: As pointed out by one of the replies on the question, you can notice that $$ \lim_{x\to 0} \frac {ln(1+e^x)-ln2}{x} $$ is written in the format of one of the theoretical definition of derivative at point x : 
$$ \lim_{h\to 0} \frac {f(x+h) - f(x)}{x} $$
Therefore, x can be evaluated to 0 and $ f(x) = ln(1+e^x)$ 
Using this function, calculate its derivative which is $$ f'(x) = \frac {1}{1+e^x} (e^x) $$ Then you plug $ x=0 $ to get the answer $$ f'(1) = \frac{e^0}{1+e^0} = \frac{1}{2} $$