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I'm having a hard time characterising the behavior of the following expression:


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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1 Answer 1

up vote 8 down vote accepted

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 $$ K_n=\left(1-\frac{c^2}n+o\left(\frac1n\right)\right)^{n/4}\longrightarrow\exp\left(-\frac{c^2}4\right). $$

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Thanks, very nice proof! –  M.B.M. Oct 31 '11 at 22:27
I posted another limit question, which is sort of related to this one: math.stackexchange.com/questions/77655/… –  M.B.M. Oct 31 '11 at 23:05

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