Limit of a standard dissection I am looking for the limit of the following expression $$\lim_{k\rightarrow\infty}\frac{(\frac{1}{k}+\frac{\epsilon}{j})^j(\frac{1}{k}-\frac{\epsilon}{k-j})^{k-j}}{(\frac{1}{k})^k}$$
with $j<k$ and some $\epsilon>0$.
What I have done so far:
We can rewrite it as $$\lim_{k\to \infty}{(1+\frac{k\epsilon}{j})^j(1-\frac{k\epsilon}{k-j})^{k-j}}$$. I have no idea how to compute the limit of the above expression as the first term goes to infinity and the latter one to zero.(I have thought about using L'Hospital and derive the expression $j$ times until the first term has power 0, i.e. 1) I have noticed that for different value of $\epsilon$, this sequence might converge to 0 or diverge. How to find the criteria of $\epsilon$, such that this sequence converges?
 A: Note that for $k\to +\infty$


*

*$(\frac{1}{k}+\frac{\epsilon}{j})^j =\left(\frac{1}{k}\right)^{j}(1+\frac{\epsilon k}{j})^j $

*$(\frac{1}{k}-\frac{\epsilon}{k-j})^{k-j}=\left(\frac{1}{k}\right)^{k-j}(1-\frac{\epsilon k}{k-j})^{k-j}$


thus
$$\frac{(\frac{1}{k}+\frac{\epsilon}{j})^j (\frac{1}{k}-\frac{\epsilon}{k-j})^{k-j}}{(\frac{1}{k})^k}
=\frac{\left(\frac{1}{k}\right)^{j}(1+\frac{\epsilon k}{j})^j  \left(\frac{1}{k}\right)^{k-j}(1-\frac{\epsilon k}{k-j})^{k-j}}{(\frac{1}{k})^k}
=\left(1+\frac{\epsilon k}{j}\right)^j  \left(1-\frac{\epsilon k}{k-j}\right)^{k-j}=e^{j\log\left(1+\frac{\epsilon k}{j}\right)+(k-j)\log{\left(1-\frac{\epsilon k}{k-j}\right)}}$$
and for $0<\epsilon<1$
$$j\log\left(1+\frac{\epsilon k}{j}\right)+(k-j)\log{\left(1-\frac{\epsilon k}{k-j}\right)}=k\left[ j\frac{\log\left(1+\frac{\epsilon k}{j}\right)}{k}+\frac{(k-j)}{k}\log{\left(1-\frac{\epsilon k}{k-j}\right)}\right]\to +\infty[0+1\cdot\log(1-\epsilon)]=-\infty$$
For $\epsilon>1$
$$\left(1-\frac{\epsilon k}{k-j}\right)>-1$$
thus
$$\left(1+\frac{\epsilon k}{j}\right)^j  \left(1-\frac{\epsilon k}{k-j}\right)^{k-j}\to +\infty\cdot \pm\infty=\pm\infty$$
For $\epsilon=1$
$$\left(1+\frac{k}{j}\right)^j  \left(1-\frac{k}{k-j}\right)^{k-j}=\left(1+\frac{k}{j}\right)^j  \left(\frac{-j}{k-j}\right)^{k-j}=\left(\frac{j+k}{j}\right)^j \left(\frac{j-k}{j}\right)^{j} \left(\frac{j}{j-k}\right)^{k}=\frac{1}{j^{2j}}(j^2-k^2)^j\left(\frac{j}{j-k}\right)^{k}$$
by ratio test of the absolute value we obtain
$$\frac{((k+1)^2-j^2)^j\left(\frac{j}{k+1-j}\right)^{k+1}}{(k^2-j^2)^j\left(\frac{j}{k-j}\right)^{k}}
=\frac{((k+1)^2-j^2)^j}{(k^2-j^2)^j} \left(\frac{j}{k+1-j}\right) {\left(\frac{k-j}{k+1-j}\right)^{k}}
=\frac{((k+1)^2-j^2)^j}{(k^2-j^2)^j} \left(\frac{j}{k+1-j}\right) {\left(1-\frac{1}{k+1-j}\right)^{k}}
=\frac{((k+1)^2-j^2)^j}{(k^2-j^2)^j} \left(\frac{j}{k+1-j}\right) {\left(1-\frac{1}{k+1-j}\right)^{j-1}} {\left(1-\frac{1}{k+1-j}\right)^{k+1-j}}\\\to 1\cdot 0\cdot 1 \cdot \frac1e=0$$

Therefore 
$$\lim_{k\rightarrow\infty} \frac{(\frac{1}{k}+\frac{\epsilon}{j})^j (\frac{1}{k}-\frac{\epsilon}{k-j})^{k-j}}{(\frac{1}{k})^k}=
\begin{cases}
0\quad \text{for} \quad 0<\epsilon\le1\\
\\
\pm\infty \quad \text{for} \quad \epsilon>1\end{cases}$$

A: I'm assuming $\epsilon, j$ are fixed. I'll use $f_k$ to denote the expression in question. Note that $$f_k = \left( \frac{ k\epsilon + j}{j} \right)^j \left( \frac{- (k(\epsilon - 1) + j)}{k-j} \right)^{k-j}.$$

First let $\epsilon < 1$
Let $k$ be large enough so that  $k \epsilon < k-j \iff \epsilon < 1- j/k,$ and $k - j > 1$. Using $(1 - x/n)^n < e^{-x}$ for $|x| < n, n>1,$ we have that for large $k,$ \begin{align*}
0 < \left(1 + \frac{k\epsilon}{j} \right)^j \left(1 - \frac{k\epsilon}{k-j} \right)^{k-j} \le \left(1 + \frac{k\epsilon}{j} \right)^j e^{-k\epsilon} = O(k^j) e^{-k \epsilon}, 
\end{align*}
and this upper bound goes to $0$. Since for any fixed $j,$ $1- j/k \to 1,$ we thus have that for $\epsilon < 1,$ the expression goes to $0$.

Now let $\epsilon > 1.$ Note that
$$ |f_k| = \frac{(k \epsilon + j)^j }{j^j} \left(\frac{k (\epsilon - 1) + j}{k-j}\right)^{k-j} > \left(\frac{k\epsilon}{j}\right)^j (\epsilon - 1)^{k-1},$$
and thus $|f_k|$ diverges. In particular, the sign of $f_k$ will flip about, and thus the limit points of $f_k$ are $\pm \infty$.

For $\epsilon = 1,$  \begin{align*}
|f_k| &= \frac{k^j ( 1 + j/k)^j}{j^j} \frac{j^{k-j}}{k^{k-j} (1- j/k)^{k-j}} \\ &= \left( \frac{j}{k} \right)^{k - 2j} \frac{(1 + j/k)^j}{(1 - j/k)^{k-j}} \end{align*}
Now, as $k \uparrow \infty,$ $(1 + j/k)^j \to 1, (1- j/k)^{k-j} \to e^{-j} > 0,$  and $(j/k)^{k - 2j} \to 0,$ all of which are finite. Thus, as $k \uparrow \infty, f_k \to 0.$
