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


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


with the following constraints on the parameters: $0<b<a<\infty$, 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. $f(n)$ vs. $n$ for $\epsilon=0.3$ $f(n)$ 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!

  • $\begingroup$ Have you tried re-expressing it with logarithms, then using l'Hopital? $\endgroup$ – Shaun Ault Oct 31 '11 at 23:39
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    $\begingroup$ Hmmm... helpful comment in that it got me thinking. and I think I can prove that it goes to zero! $$\begin{align}\left(\frac{2\sqrt{a(a+b/n^{0.5-\epsilon})}}{2a+b/n^{0.5-\epsilon}}\right)^{\frac{n}{2}}&=&\left(\frac{4a^2+4ab/n^{0.5-\epsilon}}{(2a+b/n^{0.5-\epsilon})^2}\right)^{\frac{n}{4}}\\ &=&\left(\frac{4a^2+4ab/n^{0.5-\epsilon}}{4a^2+4ab/n^{0.5-\epsilon}+b^2/n^{1-2\epsilon}}\right)^{\frac{n}{4}} \end{align}$$ Since the expression inside the parens is less than one, the limit goes to zero. $\endgroup$ – M.B.M. Nov 1 '11 at 0:01
  • $\begingroup$ Bullmoose, please write that as an answer so that users get a chance to upvote. You can even accept your own answer by the way. =) (Also this question will show up as answered in the future.) $\endgroup$ – Srivatsan Nov 1 '11 at 2:02
  • $\begingroup$ @Srivatsan: I am now not sure of my answer... I will update the question. $\endgroup$ – M.B.M. Nov 1 '11 at 5:34
  • $\begingroup$ Since the expression inside the parens is less than one, the limit goes to zero... Hmmgg, that hurted. Compare $(1-1/n^2)^n$, $(1-1/n)^n$ and $(1-1/\sqrt{n})^n$. $\endgroup$ – Did Nov 1 '11 at 8:18

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 $\epsilon\geqslant1/2$.

  • $\begingroup$ This is beautiful. Thank you! $\endgroup$ – M.B.M. Nov 1 '11 at 17:21

if you make substitution:

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

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

  • $\begingroup$ Hmmm... interesting approach. I see nothing wrong with it mathematically at the moment, except that it is not consistent with this answer. Can you explain the reason why they are different for $\epsilon=0$ (they are identical at $\epsilon=0$ in both expressions)? Also, how can one explain the puzzling behavior of the function whose limit is being taken shown on the figures I have uploaded just now... $\endgroup$ – M.B.M. Nov 1 '11 at 7:15
  • $\begingroup$ I really don't know why Matlab graphs looks like that...I think that limit should be irrespective of $\epsilon$ value if$-0.5<\epsilon<0.5$ $\endgroup$ – Peđa Terzić Nov 1 '11 at 7:37
  • $\begingroup$ How about the inconsistency with the answer to this question? $\endgroup$ – M.B.M. Nov 1 '11 at 7:46
  • $\begingroup$ which part of my answer is incorrect ? $\endgroup$ – Peđa Terzić Nov 1 '11 at 8:50
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    $\begingroup$ @pedja, the interesting thing is the extent of the misuses of, and misconceptions about, this mathematical software (and some others which are similar). Some people truly seem to have left their brain at the door in these affairs... For starters, you might want to read this. $\endgroup$ – Did Nov 1 '11 at 10:47

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