Show that $\lim_{x\to \infty}\left( 1-\frac{\lambda}{x} \right)^x = e^{-\lambda}$ Based on the definition of $e: = \lim_{x\to\infty} \left(1+\frac1x \right)^x$, how can we show that 
$$\lim_{x\to \infty}\left( 1-\frac{\lambda}{x} \right)^x  = e^{-\lambda}?$$
So far I've tried changing variables, $\eta = \frac{-x}{\lambda}$, so $=\lim_{\eta \to -\infty}\left( \left( 1 + \frac1\eta \right)^\eta \right)^{-\lambda}$. But then we would need to show $\lim_{\eta \to -\infty}\left( 1 + \frac1\eta \right)^\eta =e$.
 A: Hint:
$$\left(1+\frac1{\eta}\right)^\eta = \left(\frac1{\left(1+\frac1{-(\eta+1)}\right)^{-(\eta+1)}}\right)^{-\eta/(\eta+1)}$$
and $-(\eta+1) \to +\infty$ as $\eta \to -\infty$.
A: $\newcommand{\abs}[1]{\left\vert #1\right\vert}$
When $\lambda = 0$, the result is trivially true: $1 = {\rm e}^{0}$. Let's consider the cases $\lambda \not= 0$:


*

*$\large\lambda < 0$
$$
\lim_{x\to \infty}\left(1 - {\lambda \over x}\right)^{x}
=
\lim_{x\to \infty}\left[%
\left(1 + {\abs{\lambda} \over x}\right)^{x/\abs{\lambda}}\right]^{\abs{\lambda}}
=
\lim_{x\to \infty}\left[%
\left(1 + {1 \over x}\right)^{x}\,\right]^{\abs{\lambda}}
=
{\rm e}^{\abs{\lambda}}
=
{\rm e}^{-\lambda}
$$


*$\large\lambda > 0$
$$
\lim_{x\to \infty}\left(1 - {\lambda \over x}\right)^{x}
=
\lim_{x\to \infty}\left[%
\left(1 - {1 \over x/\lambda}\right)^{-x/\lambda}\right]^{-\lambda}
=
\lim_{x\to \infty}\left[%
\left(1 - {1 \over x}\right)^{-x}\right]^{-\lambda}
=
{\rm e}^{-\lambda}
$$



Otherwise,
$$
\lim_{x\to \infty}\left(1 - {\lambda \over x}\right)^{x}
=
\lim_{x\to \infty}
\exp\left(\vphantom{\LARGE A^{A}}\,x\ln\left(1 - {\lambda \over x}\right)\right)
=
\lim_{x\to \infty}
\exp\left(\vphantom{\LARGE A^{A}}\,x\left[-{\lambda \over x}\right]\right)
=
{\rm e}^{-\lambda}
$$
