Limit of $\binom{n}{k}/n^k$

Question

How to prove that, $n$ approach infinity and $0\leq k\leq n$ : $$ \lim_{n\to\infty} \dfrac{1}{k!}\times\dfrac{n!}{(n-k)!\times n^k} = \dfrac{1}{k!}$$

I have no idea to begin the proof...

Thanks in advance for help.

Answer 1

$$\frac{n!}{(n-k)!} = n(n-1)\cdots(n-(k+1))$$ is a polynomial in $n$ of degree $k$. So you have the limit towards infinity of a rational function whose numerator and denominator have the same degree. Thus the limit is the ratio of the leading coefficients.

$$\lim\limits_{n\to\infty} \frac{\left(\frac{n!}{(n-k)!}\right)}{n^k} = 1$$

Answer 2

Hint:

$$\lim_{n\to\infty}\dfrac n{n-k}=?$$

Answer 3

Quite an overkill, but since $n!=\Gamma(n+1)$, your limit is a consequence of Gautschi's inequality.