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I am having trouble expressing the behavior of the following limit:

$\lim_{n\rightarrow\infty}\left(\frac{2\sqrt{a(a+b/n^{0.5-\epsilon})}}{2a+b/n^{0.5-\epsilon}}\right)^{\frac{n}{2}}$

After some simple arithmetic manipulations I can simplify this expression to this:

$\lim_{n\rightarrow\infty}\left(1+\frac{b^2n^{-1+2\epsilon}}{4a^2+4abn^{-0.5+\epsilon}}\right)^{-\frac{n}{4}}$

with the following constraints on the parameters: $0, and $-0.5<\epsilon<0.5$. For $0<\epsilon<0.5$ it seems to go to zero, and for $-0.5<\epsilon<0$ it seems to go to one (at least it looks that way when plotting it in MATLAB, see pictures for $a=1$, $b=0.1$.) At $\epsilon=0.5$ it's a constant function of $a$ and $b$, according to an answer to my previous and related question. <span class=f(n) vs. $n$ for $\epsilon=0.3$"> <span class=f(n) vs. $n$ for $\epsilon=-0.1$">

I am perplexed on how to actually prove the statements for $0<\epsilon<0.5$ and $-0.5<\epsilon<0$. It'd be great if someone could help!

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    @Bullmoose, one cannot downvote comments. But who thinks that up and downvotes mean anything on MSE...2011-11-01

2 Answers 2

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Let $c=b/(2a)$. The limit is $1$ if $\epsilon<0$, $\mathrm e^{-c^2/4}$ if $\epsilon=0$, and $0$ if $\epsilon>0$.

To see this, start from the last displayed expression (right after After some simple arithmetic manipulations) and write its logarithm as $(-n/4)\log(1+c^2y_n)$ with $ y_n=\frac{n^{-1+2\epsilon}}{1+2cn^{\epsilon-1/2}}. $ First assume $\epsilon<1/2$. The denominator of $y_n$ goes to $1$, hence $y_n\sim n^{-1+2\epsilon}$. In particular $y_n\to0$, hence $\log(1+c^2y_n)\sim c^2y_n$ and $n\log(1+c^2y_n)\sim c^2ny_n\sim c^2n^{2\epsilon}$. This proves that the limit of $n\log(1+c^2y_n)$ is $+\infty$ for every $0<\epsilon<1/2$, $c^2$ for $\epsilon=0$, and $0$ for every $\epsilon<0$, which proves the desired result for every $\epsilon<1/2$.

If $\epsilon=1/2$, $y_n=1/(1+2c)$ for every $n$. If $\epsilon>1/2$, $y_n\to+\infty$. Thus, in both cases, $n\log(1+c^2y_n)\to+\infty$, which completes the proof for every $\epsilon\geqslant1/2$.

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    This is beautiful. Thank you!2011-11-01
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if you make substitution:

$t=a+\frac{b}{n^{0.5-\epsilon}}$ your limit becomes :

$\lim_{t\rightarrow a}\left(\frac{2\sqrt{at}}{a+t}\right)^{\sqrt [0.5-\epsilon]{\frac{b}{4(t-a)}}} ={{\rm e}^{\lim _{t\rightarrow a} \left(( {\frac {b}{4\,t-4\,a}} \right) ^{ \left( 0.5-\epsilon \right) ^{-1}}\cdot\ln \left( 2\,{\frac { \sqrt{at}}{a+t}} )\right) }}=e^{0}=1 $

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    Thank you for the effort on your attempt at the answer. I suspected that something was wrong from the MATLAB graphs which I posted. @DidierPiau might not like that approach entirely, but it did confirm [his solution](http://math.stackexchange.com/questions/77629/behavior-of-lim-n-rightarrow-infty-left-frac2-sqrtaab-sqrtn-epsilo) previously, for this particular family of functions. I grew further suspicious this morning when I realised that one can "show" that $\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n=1$ using substitution $m=1/n$ and similar logic...2011-11-01