Evaluate this limit involving e I am solving this limit, but cannot find a satisfactory solution:
\begin{align}
\lim_{\alpha\to\infty}\frac{\ln(1+e^\alpha)}{\alpha} 
\end{align}
I have tried substitution like $y = e^\alpha$ and $y = 1+ e^\alpha$ but nothing seems to work.  
 A: $$\lim_{\alpha \to \infty} \frac{\ln(1+e^{\alpha})}{\alpha} \overset{\text{L.H.}}= \lim_{\alpha \to \infty}\frac{e^{\alpha}}{1+e^{\alpha}} $$ Dividing by $e^{\alpha}$ gives $$\lim_{\alpha \to \infty} \frac{1}{e^{-\alpha} + 1}=\frac{1}{0+1} =1 $$
A: Hint:
Use the Squeeze Theorem:

$$\lim_{\alpha\rightarrow\infty}\frac{\ln(e^\alpha)}{\alpha}\leq\lim_{\alpha\rightarrow\infty}\frac{\ln(1+e^\alpha)}{\alpha}\leq \lim_{\alpha\rightarrow\infty}\frac{\ln(e^\alpha+e^\alpha)}{\alpha}\\\ln(e)\leq \lim_{\alpha\rightarrow\infty}\frac{\ln(1+e^\alpha)}{\alpha}\leq \ln(e)+\lim_{\alpha\rightarrow \infty}\frac{\ln(2)}{\alpha}$$

A: Note that
$$\ln(1+e^a)=\ln(e^a(e^{-a}+1))=\ln(e^a)+\ln(e^{-a}+1)=a+\ln(e^{-a}+1)$$
Thus
$${\ln(1+e^a)\over a}=1+{\ln(e^{-a}+1)\over a}\to1+{\ln(0+1)\over\infty}=1+{0\over\infty}=1$$
A: Quite short, with some asymptotic analysis:
$1+\mathrm e^α\sim_\infty \mathrm e^α,\,$ so $\,\ln\bigl(1+\mathrm e^α\bigr)\sim_\infty \ln\bigl(\mathrm e^α\bigr)=α$, and therefore
$$\frac{\ln\bigl(1+\mathrm e^α\bigr)}{α}\sim_\infty\frac{α}{α}=1.$$
