I am trying to evaluate the following limit:
$$
\lim_{n->\infty}\frac{(n-k)!}{n!}
$$
where $k$ is a non-negative integer.
I tried two different appraoches. Strangely, I am getting 2 different answers.
Approach 1:
$$
\lim_{n->\infty}\frac{(n-k)!}{n!} = \lim_{n->\infty}\frac{n!}{n!} = 1
$$
In approach 1, I am thinking that $(n-k)!$ will become $n!$ when $n$ is large (approaches infinity).
Approach 2:
$$
\lim_{n->\infty}\frac{(n-k)!}{n!} = \lim_{n->\infty}\frac{n^{n-k} + O(n^{n-k-1})}{n^n+O(n^{n-1})} = \frac{1}{n^{k}}
$$
Clearly, $ \frac{1}{n^{k}} \ne 1$ for $k \gt 0$.
Why am I getting different answers from the 2 methods above?
Best Answer
What I see is that you are wrongly interpreting the meaning of a constant in this limit.
First of all, What is really the answer to this question?
As already pointed out in the comments,
Case 1: $k=0$.
$lim_{n\to\infty} \frac{(n-k)!}{n!} = lim_{n\to\infty} \frac{n!}{n!} = 1$.
Case 2: $k \geq 1$.
$lim_{n\to\infty} \frac{(n-k)!}{n!} = lim_{n\to\infty} \frac{1}{(n-k+1)(n-k+2)\cdots(n)} = 0$
Now for the intuition part, Why is this case that even though k is just a constant, still, the limit depends on k?
For that, what you can do is get your calculator out and try the the same problem but with (ofcourse) finite numbers (increasing them along the way) and you will observe that limit indeed tends to zero $\forall \ k \geq 1$ because what’s left in the denominator always $\to \infty$