Binomial limit $p = \lambda/n, n \to \infty,\ Pr(X= k)$ $n = n$
$p = \lambda/n$
Binomialpdf evaluated at $x = k$,
What is it's limit as $n$ tends to infinity?
The answer should contain only variables $\lambda$ and $k$. I'm hopeless at limits - could someone please show me how to solve this one?
 A: Let's write it out:
$$P(X=k) = \binom{n}{k}\cdot p^k\cdot (1-p)^{n-k}$$
$$P_n(X=k) = \frac{n(n-1)(n-2)\cdots (n-k+1)}{1\cdot 2\cdot 3\cdots k}\cdot \left(\frac{\lambda}{n}\right)^k\cdot \left(1-\frac{\lambda}{n}\right)^{n-k}$$
$$P_n(X=k) = \frac{n(n-1)(n-2)\cdots (n-k+1)}{n\cdot n\cdot n\cdots n}\cdot \frac{\lambda^k}{k!}\cdot \left(1-\frac{\lambda}{n}\right)^{n-k}$$
Now, we take the limit as $n\to\infty$. In that first fraction, each of the $n$ terms of the form $\frac{n-j}{n}$ goes to $1$, so their product (that fraction) goes to $1$. The next fraction $\frac{\lambda^k}{k!}$ is constant, depending only on the variables we're keeping - we leave that alone. Finally, there's that last term, a $1^{\infty}$ indeterminate form. Its logarithm is
$$\ln\left(1-\frac{\lambda}{n}\right)^{n-k} = (n-k)\ln\left(1-\frac{\lambda}{n}\right)=\frac{\lambda}{p}\ln(1-p)-k\ln(1-p)$$
Now, we have the well-known limit $\lim_{t\to 0}\frac{\ln(1-t)}{t}=-1$, so $\lim_{p\to 0}\frac{\lambda}{p}\ln(1-p)-k\ln(1-p)=-\lambda$. Transforming back to $n$ instead of $p$ and taking an exponential to undo the logarithm,
$$\lim_{n\to\infty}\left(1-\frac{\lambda}{n}\right)^{n-k} = e^{-\lambda}$$
Now, we put the pieces together. The product of our three limits gives us
$$\lim_{n\to\infty}P_n(X=k) = \frac{\lambda^k}{k!}e^{-\lambda}$$
This is the well-known and broadly useful Poisson distribution.
