Find the limit of $\lim_{x\to0}{\frac{\ln(1+e^x)-\ln2}{x}}$ without L'Hospital's rule

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.

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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 $x<1$

$$1+x\le e^x\le \frac{1}{1-x}\tag2$$

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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.

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 $f$!!

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 $z = y -1$

$$L = \lim_{z\to0}{\frac{\ln(z + 2)-\ln2}{\ln(z+1)}} = \dfrac12\lim_{z\to0}\frac{\ln(z/2 + 1)}{z/2}\dfrac{z}{\ln(z+1)}= \dfrac12$$