Dùng cách hơi trâu bò tí...
Ta có $\lim_{x\to0} \left(\frac{\ln(\sqrt[3]{3x+1}+1)}x-\frac{\ln(\sqrt{x+1}+1)}{x}\right)$
$=\lim_{x\to0} \left( \dfrac{\ln \left(\dfrac{\sqrt[3]{3x+1}+1}{\sqrt{x+1}+1}\right)}{x}\right)$
$=\lim_{x\to0}\left( \frac{\ln\left(\dfrac {\sqrt[3]{3x+1}-\sqrt{x+1}}{\sqrt{x+1}+1}+1\right)}{\dfrac {\sqrt[3]{3x+1}-\sqrt{x+1}}{\sqrt{x+1}+1}}.\frac{\dfrac {\sqrt[3]{3x+1}-\sqrt{x+1}}{\sqrt{x+1}+1}}{x}\right)$
$=1.\lim_{x\to0}\frac{\sqrt[3]{3x+1}-\sqrt{x+1}}{x(\sqrt{x+1}+1)} $
$=\lim_{x\to0} \frac{\sqrt[3]{(3x+1)^2}-(x+1)}{x(\sqrt{x+1}+1)(\sqrt[3]{3x+1}+\sqrt{x+1})}$
$=\lim_{x\to0} \frac{(3x+1)^2-(x+1)^3}{x.\color{green}{f(x)}}=\lim_{x\to0} \frac{x(-x^2+6x+3)}{x \color{green}{f(x)}}$
$=\lim_{x \to 0}\frac{-x^2+6x+3}{\color{green}{f(x)}}$
Trong đó $\color{green}{f(x)}=(\sqrt{x+1}+1)(\sqrt[3]{3x+1}+\sqrt{x+1})\left[\sqrt[3]{(3x+1)^2}+\sqrt[3]{(3x+1)^2}(x+1)+(x+1)^2\right]$
Và $\lim_{x\to0}\color{green}{f(x)}=12$
Do đó giới hạn cần tìm là $\frac 14$