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

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