Help clarify the limit of $e$ when there is an exponent I want to compute:
$$\lim_{n\to \infty} \left(1-\frac{\lambda}{n}\right)^n$$
I know that:
$$e = \lim_{x\to \infty}\left(1+\frac{1}{x}\right)^x$$
So, I let $x=-\frac{n}{\lambda}$ and get:
$$\lim_{n\to \infty} \left(1-\frac{\lambda}{n}\right) = \lim_{n\to \infty} \left(1+\frac{1}{x}\right)^{x(-\lambda)} = e^{-\lambda}$$
My question is how do I justify going from $\lim_{n\to \infty} \left(1+\frac{1}{x}\right)^{x(-\lambda)}$ to $e^{-\lambda}$?
I thought of the following:
$$\lim_{n\to \infty} \left(1+\frac{1}{x}\right)^{x(-\lambda)}=\left[\lim_{n\to \infty} \left(1+\frac{1}{x}\right)^x\right]^{-\lambda} = \left[ e \right]^{-\lambda} = e^{-\lambda}$$
I don't think that is right since $-\lambda$ is part of the expression that I want to find the limit for and I don't recall ever learning that we can factor out an exponent from a limit expression.
Please help.
 A: We deal with the interesting case $\lambda$ positive. Note that 
$$1-\frac{\lambda}{n}=\frac{1}{1+\frac{\lambda}{n-\lambda}}$$
(unless $n=\lambda$). Take the $n$-th power of the expression on the right. To make typing easier, study only the $n$-th power of the denominator. 
Let $x=\frac{n-\lambda}{\lambda}$. Then $n=\lambda+x\lambda$. Thus the $n$-th power of the denominator is 
$$\left(1+\frac{1}{x}\right)^{\lambda+x\lambda}.$$
This can be written as 
$$\left(1+\frac{1}{x}\right)^{\lambda}\left(\left(1+\frac{1}{x}\right)^{x}\right)^\lambda.\tag{1}$$
Now the limit process is uneventful. As $n\to\infty$, $x\to\infty$. Since the function $t^\lambda$ is continuous at $t=e$, the limit of (1) is $e^\lambda$, and therefore our sought-for limit is $e^{-\lambda}$. 
A: $$\begin{align}\lim_{n\to \infty} \left(1-\frac{\lambda}{n}\right)^n
&=\lim_{n\to \infty} \left(\frac{n}{n-\lambda}\right)^{-n}=\lim_{n\to \infty} \left(1+\frac{\lambda}{n-\lambda}\right)^{-n}=\\
&=\lim_{n\to\infty} \left(1+\frac{1}{n/\lambda-1}\right)^{(n/\lambda-1)(-\lambda)-\lambda}=\\
&=\underbrace{\left(\lim_{n\to \infty} \left(1+\frac{1}{n/\lambda-1}\right)^{n/\lambda-1}\right)^{-\lambda}}_{e^{-\lambda}}\cdot \underbrace{\lim_{n\to\infty} \left(1+\frac{1}{n/\lambda-1}\right)^{-\lambda}}_1\end{align}$$
