Limit of $\frac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}}$

Question

I'm trying to calculate the limit of $$\frac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}}$$ as $n \to \infty$. I know (using Wolfram Alpha) that the limit should be equal to $e^{k-1}$. However I'm unable to manipulate the expression in the way that I could see the limit to be true.

Answer 1

Just split it up appropriately. A possible way is as follows:

$$\frac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}} = \frac{(n-1)^{2n}\cdot (n-1-k)^{k+1}}{n^n\cdot (n-1-k)^n \cdot (n-1)^{k+1}} $$ $$ = \underbrace{\frac{(n-1-k)^{k+1}}{(n-1)^{k+1}}}_{\stackrel{n \to \infty}{\longrightarrow}1}\cdot \underbrace{\frac{(n-1)^n}{n^n}}_{\stackrel{n \to \infty}{\longrightarrow}e^{-1}}\cdot \underbrace{\frac{1}{\frac{(n-1-k)^n}{(n-1)^n}}}_{\stackrel{n \to \infty}{\longrightarrow}\frac{1}{e^{-k}} =e^{k}}$$

Answer 2

HINT

We have

$$\dfrac{(n-1)^{2n-1-k}}{n^n (n-1-k)^{n-1-k}}=\dfrac{(n-1)^n}{n^n }\dfrac{(n-1)^{n-1-k}}{ (n-1-k)^{n-1-k}}=\left( 1-\frac1n \right)^n\frac{1}{\left( 1-\frac k{n-1} \right)^{n-1-k}}$$