Limits

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Introduction

This is some reference material on various limits.

\;x \cdot \alpha^x

\; \equiv \ln x = \int_{1}^{x} \frac{1}{t}dt


Choose \;c > 0. For \;t \ge 1 we have \;t^c \ge 1 and \;\frac{1}{t} \le \frac{1}{t} \cdot t^c = t^{c-1}.


Now for any \;a > 0, b > 0, x > 1:


\;0 \;<\; \ln x \; = \; \int_1^x \frac{1}{t} dt \; \le \; \int_1^x t^{c-1} dt \;=\; \frac{t^c}{c}\biggr\vert_1^x \;=\; \frac{x^c-1}{c} \;<\; \frac{x^c}{c}.


i.e. \;0 < \ln x \;<\; \frac{x^c}{c} for \;c>0. Now raise everything to the \;b power and divide by \;x^a where \;x^a > 0.


\;\frac{0^b}{x^a} < \frac{(\ln x)^b}{x^a} < \frac{x^{bc}}{c^b} \cdot \frac{1}{x^a} = \frac{x^{bc-a}}{c^b} (again: \;c > 0)


Take \;c = a/2b > 0:


\;0 < \frac{(\ln x)^b}{x^a} < \frac{x^{-a/2}}{(a/2b)^b} = \frac{1}{(a/2b)^b \; x^{a/2}}.


\;\frac{1}{(a/2b)^b} \lim_{x \rightarrow\infty} \frac{1}{x^{a/2}} \rightarrow 0 so \;\lim_{x \rightarrow\infty} \frac{(\ln x)^b}{x^a} \rightarrow 0.


Now setting \;K = \left(\frac{a}{2b}\right)^b we have


\;0 < \frac{(\ln x)^b}{x^a} < \frac{x^{-a/2}}{K} for all \;a, b > 0. Let \;y = \ln x so that \;e^y = x.


\;0 < \frac{y^b}{e^{ay}} < \frac{e^{-ay/2}}{K}


so \;\frac{y^b}{e^{ay}} \rightarrow 0 as \;y \rightarrow \infty.


Here of course we are really combining the idea that \;x \rightarrow \infty and \;\ln x is unbounded so it's ok to let \;y go to infinity as \;x goes to infinity.


We can put this into a different form involving an arbitrary number between 0 and 1 by taking a special case where \;b = 1, a = -\ln \alpha for some \;0 < \alpha < 1 so as above \;a > 0. Then


\;\frac{y^b}{e^{ay}} = \frac{y}{(e^a)^y} = \frac{y}{(e^{-\ln \alpha})^y} = y\;\left(e^{\ln \alpha}\right)^y = y\;\alpha^y \rightarrow 0 as \;y \rightarrow\infty.


In words any constant between zero and one raised to the y power goes to zero harder than y (raised to any power) goes to infinity, so their product in that limit goes to zero. In poker terminology: Exponentials beat powers which in turn beat straights.



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