Finding $\displaystyle \lim_{x\to 0} \frac{\ln(3^{x}+1)-\ln(2)}{x}$ I have to find the limit of $\displaystyle \lim_{x\to 0} \frac{\ln(3^{x}+1)-\ln(2)}{x}$ using only notable limits avoiding l'Hopital's method and derivatives.
I tried to use logarithm's properties but then I came in a bad situation. Then I tried to use +1 and -1 and x/x in the argument of the logarithm in order to use the notable limit so I'm stuck here... Can somebody help me in a clear way please? (tomorrow I have the limit's test!)
 A: You could use the two standard limits
$$\lim_{x\to0}\frac{\ln(x+1)}{x}=1$$
and
$$\lim_{x\to0}\frac{e^{x}-1}{x}=1.$$
We have
$$
\lim_{x\to0}\frac{\ln(3^x+1)-\ln2}{x}=
\lim_{x\to0}\frac{\ln(3^x-1+2)-\ln2}{x}=
\lim_{x\to0}\frac{\ln\left(\frac{3^x-1}{2}+1\right)+\ln2-\ln2}{x}=
\lim_{x\to0}\frac{\ln\left(\frac{3^x-1}{2}+1\right)}{\frac{3^x-1}{2}}\cdot\frac{\frac{3^x-1}{2}}{x}=
\lim_{x\to0}\frac{\ln3}{2}\cdot\frac{\ln\left(\frac{3^x-1}{2}+1\right)}{\frac{3^x-1}{2}}\cdot\frac{e^{x\ln3}-1}{x\ln3}=\frac{\ln3}{2}.
$$
A: Hints:
$$\begin{align*}\bullet&\;\;(\log x)'=\frac1x\;\;\text{, and for a differentiable function}\;\;f(x)\;,\;\;(\log(f(x))'=\frac{f'(x)}{f(x)}\\{}\\\bullet&\;\;\lim_{x\to 0}\frac{\log(3^x+1)-\log 2}x=:(\log(3^x+1))'|_{x=0}\end{align*}$$
A: $\lim_{x\to0}\dfrac{\ln(3^x+1)-\ln2}{x}\\\left(\dfrac{0}{0}\text{ form}:f(x)=\ln(3^x+1)-\ln2,~g(x)=x\text{ are diferentiable on a neighborhood of }0,~g'(x)\ne0\text{ on a deleted neighborhood of } 0 \text{ and }f(x)\to0,~g(x)\to0~as~x\to0\right)\\=\lim_{x\to0}\dfrac{\dfrac{3^x}{3^x+1}\ln 3}{1}\\=\ln 3\dfrac{3^0}{3^0+1}\left(\text{Since }{\dfrac{3^x}{3^x+1}\ln 3}\text{ is continuous on }\mathbb R\right)\\=\dfrac{1}{2}\ln3$
