Let $\epsilon_1^k$, $\epsilon_2^k$, and $\delta_2^k$ be sequences (indexed by $k$) strictly between $0$ and $1$ and such that $\epsilon_1^k \rightarrow 0$, $\epsilon_2^k \rightarrow 0$, and $\delta_2^k \rightarrow 0$. Show that it is always possible to find sequences $\epsilon_1^k$, $\epsilon_2^k$, and $\delta_2^k$ which satisfy the aforementioned conditions and such that $\frac{(1-\epsilon_1^k)\delta_2^k}{2[\epsilon_1^k\epsilon_2^k + (1-\epsilon_1^k)\delta_2^k]} = c$ where $c$ is a constant in $(0,\frac{1}{2})$.
My intuition says that we can somehow find these 3 sequences such that they are functions of each other so that they still all converge to $0$ individually while the expression simplifies to a constant. However, I am not exactly sure how to show it formally.
Start with $\epsilon_1^k\to 0$, $\epsilon_2^k\to 0$ (and with values strictly between $0$ and $1$) and fix $c'>0$. Then we can let $$ \delta_2^k=\frac{\epsilon_1^k\epsilon_2^k}{c'(1-\epsilon_1^k)}$$ Clearly, $\delta_2^k\to 0$ and $\delta_2^k>0$ for all $k$. After removing at most finitely many terms from $\delta_2$ (and the other sequences) if necessary, we thus also may assume $\delta_2^k\in(0,1)$ for all $k$. Then $$\frac{(1-\epsilon_1^k)\delta_2^k}{2[\epsilon_1^k\epsilon_2^k+(1-\epsilon_1^k)\delta_2^k]} =\frac1{2\cdot\frac{\epsilon_1^k\epsilon_2^k+(1-\epsilon_1^k)\delta_2^k}{(1-\epsilon_1^k)\delta_2^k}}=\frac1{2\cdot(c'+1)}{}$$ is constant and $\in(0,\frac12)$. In fact, given $c\in(0,\frac12)$, we can let $c'=\frac1{2c}-1$ to achieve the given constant $c$.