Why is $\lim\limits_{n\to \infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$ not equal to $1$? So I saw the limit $\lim\limits_{n\to \infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$ here the other day:
Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
and when I saw it, I right away thought the answer is $1$ because I thought $\lim\limits_{n\to \infty} \sum_{k=0}^n \frac{n^k}{k!} = \lim\limits_{n\to \infty} e^n$ given that $e^x = \lim\limits_{n\to \infty} \sum_{k=0}^n \frac{x^k}{k!}$ and so the result would be $\lim\limits_{n\to \infty} e^{-n}e^n = 1$ but the result is $\frac{1}{2}$, found using methods that I'm not familiar with.
Could someone please explain why my method is wrong?
Thank you so much in advance!

What you're doing is taking the identity
$$
\lim_{n\to\infty}\sum_{k=0}^n\frac{x^k}{k!}=e^x\tag1
$$
and plugging in $x=n$ to obtain the (false) statement
$$
\lim_{n\to\infty}\sum_{k=0}^n\frac{n^k}{k!}=e^n.\tag2
$$
Why is (2) false? Setting $x=n$ in (1) is illegal because the $n$ in (1) is busy being used as the label for the $n$th term in your sequence; plugging $x=n$ confuses $x$ with $n$ and changes the nature of the expression you're studying. To see why (2) makes no sense, notice that the LHS of (2) should no longer depend on $n$ when you've passed to the limit, so the RHS should not depend on $n$ either. For more examples of what can go wrong, try setting $x=n$ in the following identities, which are valid for all $x$:
$$
\lim_{n\to\infty}\frac xn=0\tag3
$$
and
$$
\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^{x}\tag4
$$
