The limit $lim_{nto infty}frac{T_n(n)}{e^n}$ where $T_n(x)$ is the Taylor polynomial of $e^x$

From working on a problem I was lead to consider the function $\frac{T_n(n)}{e^n}$ where $T_n(x)$ is the $n$'th order Taylor polynomial of $e^x$.

Numerical evidence suggest that

$$\lim_{n\to \infty} \frac{T_n(n)}{e^n} \equiv\lim_{n\to \infty} \frac{\sum_{k=0}^n\frac{n^k}{k!}}{\sum_{k=0}^\infty\frac{n^k}{k!}} = \frac{1}{2}$$

Is there a nice proof for this statement? More generally: is there a 'standard' approach for evaluating limits on the form $\lim_{n\to\infty}\frac{f_n(x_n)}{f(x_n)}$ where $f_n$ is a series converging (uniformly) to $f$ and where $x_n$ is an unbounded sequence? I would also apprechiate refs. to similar questions on this site or in the literature (I could only find this one).

Answer

Look here:
A limit involves series and factorials

That answer links to here:
http://journals.cambridge.org/download.php?file=%2FPEM%2FPEM2_24_03%2FS0013091500016503a.pdf&code=fd828d6902ca6a380244640216120c97

This has a result of
(who else)

Ramanujan
where he proved
(in S. RAMANUJAN, J. Ind. Math. Soc. 3 (1911), 128; ibid. 4 (1911), 151-152; Collected Papers
(Chelsea, New York; 1962), 323-324)
that

$$e^n/2 = \sum_{k=0}^{n-1} n^k/k! + (n^n/n!) r(n)$$

where, for large $n$,
$r(n) \approx 1/3 + 4/(135n) + O(1/n^2)$.