How can I solve $\displaystyle{\lim_{x \to \infty} \frac{1}{(1+\frac{k}{x})^x}}$? How can I solve $\displaystyle{\lim_{x \to \infty} \frac{1}{(1+\frac{k}{x})^x}}$ using $\displaystyle{\lim_{x \to +\infty} {(1+\frac{1}{x})^x = e}}$? What should I do with the constant "$k$?
 A: Note that I'll be assuming $k > 0$.
Observe that
$$\begin{align}\lim_{x\to\infty}\left(1 + \dfrac{k}{x}\right)^x &= \lim_{x\to\infty}\left(1 + \dfrac{k}{x}\right)^{kx/k}\\
&\overset{y=x/k}{=} \lim_{y\to\infty}\left(1 + \dfrac{1}{y}\right)^{ky}\\
&= \lim_{y\to\infty}\left(\left(1 + \dfrac{1}{y}\right)^{y}\right)^k\\
&= \left(\lim_{y\to\infty}\left(1 + \dfrac{1}{y}\right)^{y}\right)^k\\
&= e^k
\end{align}$$
The interchange of $(\cdot)^k$ and $\lim$ was possible because the function $x\mapsto x^k$ is continuous.
The answer to your question should now be clear. (The function $x \mapsto 1/x$ is continuous on $(0, \infty)$ and thus, you should get $e^{-k}$.)

Edit: Adding the case $k \le 0$ as well.
Nothing needs to be said for $k = 0$. It is clearly $1$.  
Assume that $k < 0$. Before proceeding further, let us prove the following lemma.
Lemma. $\displaystyle\lim_{y \to -\infty}\left(1 + \dfrac{1}{y}\right)^{y} = e.$
Proof. Consider the limit 
$$L = \lim_{y\to-\infty}y\ln\left(1 + \dfrac{1}{y}\right) =  \lim_{y\to-\infty}\dfrac{\ln\left(1 + \dfrac{1}{y}\right)}{1/y}.$$
We can evaluate it using L'Hospital since the rightmost limit is of the form $0/0$. This gives us
$$L= \lim_{y\to-\infty}\dfrac{\left(1 + \dfrac{1}{y}\right)(-1/y^2)}{-1/y^2} = 1.$$
From this, it follows that the original limit was $e^L = e$.

With this lemma in place, the result for $k < 0$ follows by again making the substitution $y = x/k$ but now noting that $y \to -\infty$ instead.
Interestingly, the answer in all these cases turns out to be $\boxed{e^{-k}}$.
A: The following works for $k \in \mathbb{R}$, you just have to use $\ln(1+a)= a+O(a^2)$ for small a:
$$
\displaystyle{\lim_{x \to +\infty} \frac{1}{(1+\frac{k}{x})^x}}
= \lim_{x \to +\infty} (1+\frac{k}{x})^{-x} = \lim_{x \to +\infty}  e^{-x  \ln{ (1+\frac{k}{x})}}  = \lim_{x \to +\infty}  e^{-x     \frac{k}{x} } =    e^{-k} 
$$
