Evaluating a couple of limits
I wonder if someone could clarify these limits for me. They arise in the derivation of the poisson distribution from the binomial distribution as 
and 
The first is
![[1- (\lambda/n)]^{n} \rightarrow \mathrm{e}^{-\lambda}](http://latex.codecogs.com/png.latex?[1- (\lambda/n)]^{n} \rightarrow \mathrm{e}^{-\lambda})
as 
.
I don't see how this works. If anything I would have thought that
)^{n} = \mathrm{e}^{n \cdot \log{(1-\lambda/n)}} \to \mathrm{e}^{0} = 1 )
as .
The second is that
\frac{n!}{(n-k)!(n-\lambda)^k = \frac{n(n-1) \ldots (n-k+1)}{(n - \lambda) (n-\lambda) \ldots (n - \lambda)}
is a "quantity that tends to 1 as (since 
remains constant)."
What precisely is the role of ? Is the critical condition not that the numerator and denominator grow at the same rate, and if so, how can this be demonstrated?
Thanks in advance. MD
Re: Evaluating a couple of limits
Quote:
Originally Posted by
Mathsdog 
If anything I would have thought that
as
.
\ne0)
. Rather, using the Maclaurin series for ln, =n(-\lambda/n+o(1/n))\to-\lambda)
as 
.
Quote:
Originally Posted by
Mathsdog !(n-\lambda)^k} = \frac{n(n-1) \ldots (n-k+1)}{(n - \lambda) (n-\lambda) \ldots (n - \lambda)})
is a "quantity that tends to 1 as
(since
remains constant)."
If k is constant, then  \dots (n-k+1)}{(n - \lambda) (n-\lambda) \dots (n - \lambda)} = \lim_{n\to\infty}\frac{n}{n-\lambda} \lim_{n\to\infty}\frac{n-1}{n-\lambda}\dots\lim_{n\to\infty}\frac{n-k+1}{n-\lambda})
. Each of these limits is 1. Indeed, 
as .
Re: Evaluating a couple of limits
Quote:
Originally Posted by
Mathsdog I wonder if someone could clarify these limits for me. They arise in the derivation of the poisson distribution from the binomial distribution as
and
The first is
as
.
I don't see how this works. If anything I would have thought that
as
.
The second is that
\frac{n!}{(n-k)!(n-\lambda)^k = \frac{n(n-1) \ldots (n-k+1)}{(n - \lambda) (n-\lambda) \ldots (n - \lambda)}
is a "quantity that tends to 1 as
(since
remains constant)."
What precisely is the role of
? Is the critical condition not that the numerator and denominator grow at the same rate, and if so, how can this be demonstrated?
Thanks in advance. MD
![\displaystyle \begin{align*} \lim_{n \to \infty}\left( 1 - \frac{ \lambda}{n}\right)^n &= \lim_{n \to \infty}e^{\ln{\left[\left(1 - \frac{\lambda}{n}\right)^n\right]}} \\ &= \lim_{n \to \infty}e^{ n\ln{ \left( 1 - \frac{\lambda}{n} \right) } } \\ &= \lim_{n \to \infty} e^{ \frac{ \ln{ \left( 1 - \frac{\lambda}{n} \right) } }{ \frac{1}{n} } } \\ &= e^{\lim_{n \to \infty}\frac{\ln{ \left( 1 - \frac{\lambda}{n} \right) }}{\frac{1}{n}}} \\ &= e^{ \lim_{n \to \infty}\left(\frac{\frac{1}{n - \lambda} - \frac{1}{n}}{-\frac{1}{n^2}}\right) } \textrm{ by L'Hospital's Rule} \\ &= e^{ \lim_{n \to \infty}\left[ \frac{\frac{\lambda}{n(n + \lambda)}}{-\frac{1}{n^2}} \right] } \\ &= e^{ \lim_{n \to \infty}\left(-\frac{n^2\lambda}{n^2 + n\lambda}\right) } \\ &= e^{\lim_{n \to \infty}\left(-\frac{\lambda}{1 + \frac{\lambda}{n}}\right)} \\ &= e^{-\lambda} \end{align*}](http://latex.codecogs.com/png.latex?\displaystyle \begin{align*} \lim_{n \to \infty}\left( 1 - \frac{ \lambda}{n}\right)^n &= \lim_{n \to \infty}e^{\ln{\left[\left(1 - \frac{\lambda}{n}\right)^n\right]}} \\ &= \lim_{n \to \infty}e^{ n\ln{ \left( 1 - \frac{\lambda}{n} \right) } } \\ &= \lim_{n \to \infty} e^{ \frac{ \ln{ \left( 1 - \frac{\lambda}{n} \right) } }{ \frac{1}{n} } } \\ &= e^{\lim_{n \to \infty}\frac{\ln{ \left( 1 - \frac{\lambda}{n} \right) }}{\frac{1}{n}}} \\ &= e^{ \lim_{n \to \infty}\left(\frac{\frac{1}{n - \lambda} - \frac{1}{n}}{-\frac{1}{n^2}}\right) } \textrm{ by L'Hospital's Rule} \\ &= e^{ \lim_{n \to \infty}\left[ \frac{\frac{\lambda}{n(n + \lambda)}}{-\frac{1}{n^2}} \right] } \\ &= e^{ \lim_{n \to \infty}\left(-\frac{n^2\lambda}{n^2 + n\lambda}\right) } \\ &= e^{\lim_{n \to \infty}\left(-\frac{\lambda}{1 + \frac{\lambda}{n}}\right)} \\ &= e^{-\lambda} \end{align*})
