# Limits

## 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.