# 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:

1. 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?
2. 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 $b$?
3. what happens to this limit when $\epsilon=0$?
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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 $$K_n=\left(1-\frac{c^2}n+o\left(\frac1n\right)\right)^{n/4}\longrightarrow\exp\left(-\frac{c^2}4\right).$$
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 $L = \exp(-b^{2}/16a^{2})$.