Limit of $\left(\frac{2\sqrt{a(a+b/(\sqrt{n}+\epsilon))}}{2a+b/(\sqrt{n}+\epsilon)}\right)^{n/2}$ I'm having a hard time characterising the behavior of the following expression:
$$\lim_{n\rightarrow\infty}\left(\frac{2\sqrt{a(a+b/(\sqrt{n}+\epsilon))}}{2a+b/(\sqrt{n}+\epsilon)}\right)^{\frac{n}{2}}$$
with the following constraints on the parameters: $0<b<a<\infty$, and $\epsilon\in\mathbb{R}$.  I am interested in the following:


*

*for $\epsilon>0$, does this limit go to zero or does it go to some
constant $C$?  If it can both go to zero or to some constant $C>0$,
what are the conditions on the value of $\epsilon$ as a function of
$a$ and $b$ which leads to these outcomes, if any?

*for $\epsilon<0$, does it always go to some constant $C<1$, or can
it go to 1 for some $\epsilon$, if it's a function of $a$ and $b$?

*what happens to this limit when $\epsilon=0$?

 A: 
Claim: The limit is $\exp(-b^2/(16a^2))$, irrespective of $\epsilon$.

Proof: Let $x_n=b/(2a(\sqrt{n}+\epsilon))$, then one asks for the behaviour of 
$$
K_n=\left(\frac{1+2xx_n}{(1+xx_n)^2}\right)^{n/4}
$$
when $n\to\infty$, with $x\to0$. Note that 
$$\frac{1+2x_n}{(1+x_n)^2}=1-\frac{x_n^2}{(1+x_n)^2}=1-x_n^2+o(x_n^2),$$ 
and that $x_n^2\sim c^2/n$ with $c=b/(2a)$, hence
$$
K_n=\left(1-\frac{c^2}n+o\left(\frac1n\right)\right)^{n/4}\longrightarrow\exp\left(-\frac{c^2}4\right).
$$
A: Let the desired limit be denoted by $L$. On taking logarithms we get
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}\left(\frac{2\sqrt{a(a + b/(\sqrt{n} + \epsilon))}}{2a + b/(\sqrt{n} + \epsilon)}\right)^{n/2}\right\}\notag\\
&= \lim_{n \to \infty}\log\left(\frac{2\sqrt{a(a + b/(\sqrt{n} + \epsilon))}}{2a + b/(\sqrt{n} + \epsilon)}\right)^{n/2}\text{ (by continuity of log)}\notag\\
&= \lim_{n \to \infty}\frac{n}{2}\log\left(\frac{2\sqrt{a(a + b/(\sqrt{n} + \epsilon))}}{2a + b/(\sqrt{n} + \epsilon)}\right)\notag\\
&= \lim_{n \to \infty}\frac{n}{2}\log\left(\frac{2\sqrt{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)}}{2a(\sqrt{n} + \epsilon) + b}\right)\notag\\
&= \lim_{n \to \infty}\frac{n}{2}\log\left(1 + \frac{2\sqrt{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)} - 2a(\sqrt{n} + \epsilon) - b}{2a(\sqrt{n} + \epsilon) + b}\right)\notag\\
&= \lim_{n \to \infty}\frac{n}{2}\log(1 + (A/B))
\end{align}
where
\begin{align}
\frac{A}{B} &= \frac{2\sqrt{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)} - 2a(\sqrt{n} + \epsilon) - b}{2a(\sqrt{n} + \epsilon) + b}\notag\\
&= \frac{4\{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)\} - 4a^{2}(\sqrt{n} + \epsilon)^{2} - b^{2} - 4ab(\sqrt{n} + \epsilon)}{\left(2a(\sqrt{n} + \epsilon) + b\right)\left(2\sqrt{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)} + 2a(\sqrt{n} + \epsilon) + b\right)}\notag\\
&= -\frac{b^{2}}{\left(2a(\sqrt{n} + \epsilon) + b\right)\left(2\sqrt{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)} + 2a(\sqrt{n} + \epsilon) + b\right)}\notag\\
&\to 0 \text{ as }n \to \infty\notag
\end{align}
Therefore we can continue the limit evaluation as
\begin{align}
\log L &= \lim_{n \to 0}\frac{n}{2}\log(1 + (A/B))\notag\\
&= \lim_{n \to 0}\frac{n}{2}\cdot\frac{A}{B}\cdot\frac{\log(1 + (A/B))}{A/B}\notag\\
&= \lim_{n \to \infty}\frac{n}{2}\cdot\frac{A}{B}\notag\\
&= -\frac{b^{2}}{2}\lim_{n \to \infty}\frac{n}{\left(2a(\sqrt{n} + \epsilon) + b\right)\left(2\sqrt{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)} + 2a(\sqrt{n} + \epsilon) + b\right)}\notag\\
&= -\frac{b^{2}}{2}\lim_{n \to \infty}\frac{1}{\left(2a(1 + \epsilon/\sqrt{n}) + b/\sqrt{n}\right)\left(2\sqrt{a^{2}(1 + \epsilon/\sqrt{n})^{2} + ab(1/\sqrt{n} + \epsilon/n)} + 2a(1 + \epsilon/\sqrt{n}) + b/\sqrt{n}\right)}\notag\\
&= -\frac{b^{2}}{2}\cdot\frac{1}{2a(2a + 2a)}\notag\\
&= -\frac{b^{2}}{16a^{2}}\notag
\end{align}
Thus $L = \exp(-b^{2}/16a^{2})$.
