Find the limit of $\lim_{x\to0}{\frac{\ln(1+e^x)-\ln2}{x}}$ without L'Hospital's rule 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?
 A: Let $y = e^x$
$$L = \lim_{x\to0}{\frac{\ln(1+e^x)-\ln2}{x}} = \lim_{y\to1}{\frac{\ln(1+y)-\ln2}{\ln y}}$$
Let $z = y -1$
$$L =  \lim_{z\to0}{\frac{\ln(z + 2)-\ln2}{\ln(z+1)}} = \dfrac12\lim_{z\to0}\frac{\ln(z/2 + 1)}{z/2}\dfrac{z}{\ln(z+1)}= \dfrac12$$
A: This is just the derivative of $\ln\left(\frac{e^x+1}2\right)$ at $0$, which is indeed $\frac12$.
A: That's just the derivative of $\ln\left(e^x+1\right)$ at $0$.
If you don't see it, by Taylor's series:
$${\frac{\ln(1+e^x)-\ln2}{x}}={\frac{\ln(2+x+o(x))-\ln2}{x}}=\frac{\ln(1+\frac{x}{2}+o(x))}{x}=\frac{\frac{x}{2}+o(x)}{x}=\frac12+o(1)\to\frac12$$
A: Solution by standard limits:
set $$1+e^x=2+y \quad y\to 0\implies e^x=1+y \implies x= \ln (1+y)$$
thus
$${\frac{\ln(1+e^x)-\ln2}{x}}=\frac{\ln (2+y)-\ln 2}{\ln(1+y)}=\frac{\ln (1+\frac{y}{2})}{\ln(1+y)}=\frac{\ln (1+\frac{y}{2})}{\frac y2}\frac{y}{\ln (1+y)}\frac12\to 1\cdot 1\cdot \frac12=\frac12$$
A: There's an amusing way to prove, if the limit exists, it must be $1/2.$ Obviously, not a complete answer, since it doesn't prove the limit exists.)
Letting $f(x)=\log(1+e^x)$ then we have $f(x)=\log(1+e^{x})=\log(e^x)+\log(1+e^{-x})=x+f(-x)$ and thus $f(x)-f(-x)=x.$
So if $L=\lim_{x\to 0}\frac{f(x)-f(0)}{x}$ then $$1=\lim_{x\to 0} \frac{f(x)-f(-x)}{x}=\lim_{x\to 0}\left[\frac{f(x)-f(0)}{x}+\frac{f(-x)-f(0)}{-x}\right]=2L$$
And thus we get $L=\frac{1}{2}$.

A more complete answer uses that we know that:
$$\lim_{y\to 0}\frac{\log(1+y)}{y}=1.$$
Then we can write:
$$\frac{\log(1+e^x)-\log2}{x}=\frac{\log\left(1+\frac{e^x-1}{2}\right)}{\frac{e^x-1}{2}}\cdot \frac{\frac{e^x-1}{2}}{x}$$
Letting $y=\frac{e^x-1}{2}$, we get that $y\to 0$ as $x\to 0$ and $x=\log(1+2y)$ so we get:
$$\lim_{x\to 0}\frac{\log(1+e^x)-\log2}{x}=\lim_{y\to 0}\frac{\log(1+y)}{y}\cdot \frac{2y}{\log(2y+1)}\cdot \frac{1}{2}=1\cdot 1\cdot\frac{1}{2}$$
This is really just a long way of proving the chain rule for the derivative of $\log(e^x+1).$ 

[Complete answer updated to use the neat trick of gimusi at the end, rather than using $\frac{e^x-1}{x}\to 1$ as well.]
A: 
I thought it might be instructive to present an approach that does not rely on differential calculus, but rather uses the squeeze theorem and a set of inequalities that can be obtained with pre-calculus tools only.  To that end we proceed.


First note that in THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the logarithm and exponential functions satisfy the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1\tag 1$$
and for $x<1$
$$1+x\le e^x\le \frac{1}{1-x}\tag2$$

Next, note that $\log(1+e^x)-\log(2)=\log\left(\frac{e^x+1}2\right)$.  Hence, applying $(1)$, we can assert that
$$\frac{e^x -1}{e^x +1}\le \log(1+e^x)-\log(2)\le \frac{e^x-1}2\tag3$$
Then, applying $(2)$ to $(3)$ reveals
$$\frac{x}{e^x +1}\le \log(1+e^x)-\log(2)\le \frac{x}{2(1-x)}\tag4$$
Dividing $(4)$ by $x$, letting $x\to 0$, and applying the squeeze theorem yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{\log(1+e^x)-\log(2)}{x}=\frac12}$$
Tools Used:  The inequalities in $(1)$ and $(2)$ and the squeeze theorem.

A: 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 $$ f'(1) = \frac{e^0}{1+e^0} = \frac{1}{2} $$
A: Simply differentiate $f(x)=\ln(e^x +1)$ at the point of abscissa $x=0$ and you’ll get the answer. in fact this is the definition of the derivative of $f$!!
