[tex]\displaystyle\mit\lim_{x\to \infty}\dfrac{\ln(1+e^x)}{x}=\displaystyle\mit\lim_{x\to \infty}\dfrac{\ln \left(e^x \left(1+\dfrac{1}{e^x}\right)\right)}{x}=\displaystyle\mit\lim_{x\to \infty}\dfrac{\ln e^x+\ln \left(1+\dfrac{1}{e^x}\right)}{x}=\\ =\displaystyle\mit\lim_{x\to \infty}\dfrac{x+ \ln \left(1+\dfrac{1}{e^x}\right) }{x}= \displaystyle\mit\lim_{x\to \infty} \dfrac{x\left(1+\dfrac{\ln \left(1+\frac{1}{e^x}\right)}{x}\right)}{x} =1+\dfrac{0}{\infty}=\boxed{1}[/tex]
