[tex]l = \lim\limits_{x\to \infty}\Big(\ln(1+e^x)-x\Big) \\ \\ 1+e^x = t \Rightarrow e^x = t-1\Rightarrow x = \ln(t-1)\\ x\to \infty \Rightarrow t\to \infty\\ \\ l = \lim\limits_{t\to \infty}\Big(\ln t-\ln(t-1)\Big) = \lim\limits_{t\to \infty}\ln\Big(\dfrac{t}{t-1}\Big) =\\ \\ = \ln\Big(\lim\limits_{t\to \infty} \dfrac{t}{t-1}\Big) = \ln 1 = \boxed{0}[/tex]
