Help finding $\lim_{x \to \infty} {(1 + e^x)}^{e^{-x}}$ $\lim\limits_{x \to \infty} {(1 + e^x)}^{e^{-x}}$
Here are the steps I have taken so far : 
$\ln{L} = \lim\limits_{x \to \infty} \ln{({1 + e^x})^{e^{-x}}}\\
\ln{L} = \lim\limits_{x \to \infty} {e^{-x}} \ln{({1 + e^x)}}\\ 
\ln{L} = \lim\limits_{x \to \infty} {e^{-x}} \ln{({1 + e^x)}}\\
\ln{L} = \lim\limits_{x \to \infty} \frac{(\ln{({1 + e^x)}})}{{e^{x}}}\\
\text{Here is where I am not sure how to finish this problem. This is what I tried:}\\
\lim\limits_{x \to \infty} \frac{(\ln{({1 + e^x)}})}{{e^{x}}} = \frac{\infty}{\infty}\\
\text{Apply L'Hopitals Rule}\\
\ln{L} = \lim\limits_{x \to \infty} \frac{(\frac{1}{1+e^x} * e^x)}{e^x}\\
\ln{L} = \lim\limits_{x \to \infty} \frac{(\frac{1}{1+e^x} * e^x)}{e^x}\\
\ln{L} = \lim\limits_{x \to \infty} \require{cancel} \frac{(\frac{1}{1+e^x} * \cancel e^x)}{\cancel e^x}\\
\ln{L} = \lim\limits_{x \to \infty} \frac{1}{1+e^x}\\
\ln{L} = \lim\limits_{x \to \infty} \frac{1}{1+e^\infty} = \frac{1}{\infty} = 0 \\
\ln(0) = \text{Undefined.}$
Help is always appreciated ! also first post on stack exchange mathematics =).
 A: You are correct, except on the last one when you say $\ln L=0$. This implies that $L=1$, not that $L$ is undefined. 
A more well known result is possible. 
$e^x=t$. 
$$\lim\limits_{t \to \infty} {(1 + t)}^{\frac{1}{t}}=1$$
Or, that $$\lim \limits_{x \to o}(1+\frac{1}{x})^x=1$$
A: You only confused yourself at the very last. You have 
$$\ln L = 0$$
Taking exponential on both side, you have 
$$e^{\ln L} = e^0 = 1,$$
Thus $L = e^{\ln L}  =1$. you should get yourself familiar with this argument. After all, the fact that you can take exponential to convert $\ln L$ to $L$ is the very reason you can take $\ln$ in the first place!
A: Your procedure was correct and it's pretty clear that $\log L=0$, so that result doesn't really tell you a whole lot. The way I'd go about doing this isn't very formal, but quite intuitive. So as $x \to \infty$, it's pretty clear that $(1+e^x)$ behaves almost exactly like $e^x$. In other words, $L \sim (e^x)^{e^{-x}}=e^{xe^{-x}}$. Obviously, $xe^{-x} \to 0$ as $x \to \infty$. So $L \sim 1$ as $x \to \infty$.
A: Just another way.
Consider$$A={(1 + e^x)}^{e^{-x}}$$ Take logarithms $$\log(A)=e^{-x}\log(1+e^x)$$ Since $x$ is large $\log(1+e^x)\approx\log(e^x)=x$ So $\log(A)\approx x e^{-x}\to 0$ and then $A\to 1$.
A: By L'Hospital's Rule, $\displaystyle 0=\lim_{x \to\infty}\frac{1}{1+e^x}=\lim_{x \to\infty}\frac{\frac{e^x}{1+e^x}}{e^x}=\lim_{x \to\infty}\frac{\ln(1+e^x)}{e^x} = \lim_{x \to\infty}\ln((1+e^x)^{e^{-x}}).$
Hence, $\displaystyle\lim_{x \to\infty}(1+e^x)^{e^{-x}} = e^0=1.$
