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!

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
$$\lim_{k\to \infty} \lim_{j \to 0}\dfrac{k^j}{j! e^k} = \lim_{k\to \infty} \dfrac1{e^k} = 0$$
