calculus - Why is $limlimits_{nto infty} e^{-n}sum_{k=0}^n frac{n^k}{k!}$ 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!

Answer

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
$$\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^{x}\tag4$$