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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?

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up vote 6 down vote accepted

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 $\lim\limits_{y\to \infty}\dfrac{\ln(e^{xy}+1)}{y} = \lim\limits_{y\to \infty}\dfrac{\ln(1+1)}{y} = 0$.

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and of course if $x=0$, then $\lim_{y\to\infty}\frac{\ln(e^{xy}+1)}{y}=\lim_{y\to\infty}\frac{\ln(1+1)}{y}=0$‌​. – Hagen von Eitzen Mar 7 '13 at 16:41
@HagenvonEitzen Good point. Thanks. – Patrick Li Mar 7 '13 at 16:42

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.

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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 $\ln(e^{xy}+1)\sim\ln(e^{xy})=xy$.

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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 $$\lim\limits_{y\to +\infty}\dfrac{\ln(e^{x y}+1)}{y}=0.$$

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How do we see that the result in the first case is not valid for x<0? – vonjd Mar 7 '13 at 16:52
L'Hospital's rule is not applicable for ${x}\leqslant {0}$ . – M. Strochyk Mar 7 '13 at 17:00

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