Evaluate $\displaystyle\lim_{j\to0}\lim_{k\to\infty}\frac{k^j}{j!\,e^k}$ I found this problem in my deceased grandpa's note today when I was visiting my grandma's home.
\begin{equation}
\lim_{j\to0}\lim_{k\to\infty}\frac{k^j}{j!\,e^k}
\end{equation}
I asked my brother and he said the answer is $\cfrac{1}{2}$, but as usual, he didn't give me any explanation why the answer is $\cfrac{1}{2}$. I only get the indeterminate form $\infty^0$ in the numerator part if I substitute $j=0$ and $k=\infty$. I have no idea how to answer this problem. Could anyone here help me to answer it? I really appreciate for your help. Muchas gracias!
 A: The sum was probably transcribed wrong somewhere along the way,because it doesn't make much sense as written (the limit as $j \rightarrow 0$
of something involving $j!$ ?).  But since the answer is supposed to be $1/2$,
I'm guessing that the intended formula was something like
$$
\lim_{k \rightarrow \infty} \sum_{j=0}^k \frac{k^j} {j! e^k},
$$
which I'm sure I've seen before (possibly on this forum or on mathoverflow)
but can't easily locate a reference.  If we had replaced 
the $\sum_{j=0}^k$ by $\sum_{j=0}^\infty$ then the sum would equal $1$
for all $k$, because on writing
$$
\sum_{j=0}^\infty \frac{k^j} {j! e^k}
= \frac1{e^k} \sum_{j=0}^\infty \frac{k^j}{j!}
$$
we recognize the sum as the power series for $e^k$.
So the problem is asking in effect to prove that as $k\rightarrow\infty$
the $\sum_{j=0}^k$ part of the sum is asymptotically half of the
entire $\sum_{j=0}^\infty$ sum.  To see that this is at least plausible,
observe that the $j=k$ term
is the largest, but still accounts for only $O(1/\sqrt k)$ of the total,
and the $j=k+1, \, k+2, \, k+3, \, \ldots$ terms are approximately equal to
the $j=k-1, \, k-2, \, k-3, \, \ldots$ terms respectively.  There are
various ways to finish the proof, and your brother can probably point you
towards one of them :-)
A: Unfortunately, your brother is incorrect. Note that
$$\lim_{k\to \infty} \dfrac{k^j}{j! e^k} = 0$$
Hence,
$$\lim_{j \to 0}\lim_{k\to \infty} \dfrac{k^j}{j! e^k} = \lim_{j\to0}0 = 0$$
Even by mistake, if he had interchanged the order of the limiting process, we still get the limit as $0$, since
$$\lim_{j\to 0} \dfrac{k^j}{j! e^k} = \dfrac1{e^k}$$
and therefore
$$\lim_{k\to \infty} \lim_{j \to 0}\dfrac{k^j}{j! e^k} = \lim_{k\to \infty} \dfrac1{e^k} = 0$$
