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I have the following limit:

$$\lim_{x\rightarrow\infty}\left(1+\frac{a}{x^{1/2+\epsilon}}\left(1-\exp\left(-\frac{b}{x^{1/2+\epsilon}}\right)\right)\ln\left(\frac{a}{x^{1/2+\epsilon}}\right)\right)^x$$

where $0<b<a$.

I care for the case where $\epsilon>-1/2$. I suspect that for $\epsilon>0$ this limit evaluates to 1, and for $-1/2<\epsilon\leq0$ it evaluates to 0. However, I am having hard time evaluating this. I have tried taking the log of the expression (moving the $x$ in the exponent down), substituting $y=1/x$ and then Taylor-expanding the log, but didn't get anywhere.

Does anyone have any tips/hints that might help me evaluate this?

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Since $c=1/2+\varepsilon\gt0$, one knows that $1-\exp(-bx^{-c})\sim bx^{-c}$ and $\log(ax^{-c})\sim-c\log(x)$ when $x\to+\infty$. Thus, the function to be evaluated is $$ f(x)=(1-kg(x))^x,\quad k=abc\gt0,\quad g(x)\sim x^{-2c}\log x. $$ Since $g(x)\to0$ and $\log(1+u)\sim u$ when $u\to0$, $$ f(x)=\exp\left[x\log(1-kg(x))\right]=\exp\left[-kxg(x)\cdot(1+o(1))\right]. $$ Note that $xg(x)\sim x^{-2\varepsilon}\log x$ and recall that $k\gt0$. This yields:

  • If $-1/2\lt\varepsilon\leqslant0$, then $xg(x)\to+\infty$ hence $f(x)\to0$ when $x\to+\infty$.
  • If $\varepsilon\gt0$, then $xg(x)\to0$ hence $f(x)\to1$ when $x\to+\infty$.
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  • $\begingroup$ So it turns out I made a mistake in the original definition of the problem: I flipped the plus sign after the first 1 to a minus sign when I was typing this up (I've edited the question since I saw your answer and my mistake). However, you solved the problem that I originally had in mind (otherwise the minus in $-c\log(x)$ would've flipped the minus in your re-definition of $f(x)=(1-kg(x))^x$ to a plus). Thank you, this confirms my intuition (which was backed up by numerical evaluations)! $\endgroup$ – M.B.M. Oct 8 '12 at 5:52

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