How to prove that the limit is max(0,x)? When you do numerically experiments with the following expression it becomes pretty clear that the limit should be the max function $\max(0,x)$:
$$\lim_{y\to \infty}\frac{\ln(e^{x y}+1)}y$$
How can you prove that?
 A: If $x>0$, then $\lim\limits_{y\to \infty}\dfrac{\ln(e^{xy}+1)}{y} = \lim\limits_{y\to \infty}\dfrac{\ln(e^{xy})}{y} = x$.
If $x<0$, then $\lim\limits_{y\to \infty}\dfrac{\ln(e^{xy}+1)}{y} = \lim\limits_{y\to \infty}\dfrac{\ln(0+1)}{y} = 0$.
If $x=0$, then $\lim\limits_{y\to \infty}\dfrac{\ln(e^{xy}+1)}{y} = \lim\limits_{y\to \infty}\dfrac{\ln(1+1)}{y} = 0$.
A: Hint: Rewrite by $$\begin{align}\frac{\ln(e^{xy}+1)}{y} &= \frac{\ln\bigl(e^{xy}(1+e^{-xy})\bigr)}{y}\\ &= \frac{\ln(e^{xy})+\ln(1+e^{-xy})}{y}\\ &= \frac{xy+\ln(1+e^{-xy})}{y}\\ &= x+\frac{\ln(1+e^{-xy})}{y},\end{align}$$ to see the $x>0$ case. The $x\leq 0$ case is fairly clear as originally written.
A: Hint: The case $x\leq 0$ is clear and the limit is $0$.
Let $x>0$, then $1=_{y\to+\infty}o(e^{xy})$ so $\ln(e^{xy}+1)\sim\ln(e^{xy})=xy$. 
A: Using the L'Hospital's rule for $x>{0}$, 
$$\lim\limits_{y\to +\infty}\dfrac{\ln(e^{x y}+1)}{y}=\left( \dfrac{\infty}{\infty}\right)=\lim\limits_{y\to +\infty}\dfrac{xe^{x y}}{e^{x y}+1}=\lim\limits_{y\to +\infty}\dfrac{x}{e^{-x y}+1}=x$$
since $\lim\limits_{y\to +\infty}{e^{-x y}}=0.$
For ${x}\leqslant{0}$
$$\lim\limits_{y\to +\infty}{e^{xy}}=
\begin{cases}
1,&x=0,\\
0, &x<0,
\end{cases}
$$
therefore, in this case ${0} \leqslant \ln(e^{x y}+1)\leqslant \ln{2}$ and
$$\lim\limits_{y\to +\infty}\dfrac{\ln(e^{x y}+1)}{y}=0.$$
