GrowthOfExponentialFunction 2009-04-17 19:39:45 2009-04-17 19:39:45 Topic % this is the default PlanetPhysics preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \textbf{Lemma.} $$\lim_{x\to\infty}\frac{x^a}{e^x} = 0$$ for all \PMlinkescapetext{constant} values of $a$. {\em Proof.}\, Let $\varepsilon$ be any positive number.\, Then we get: $$0 < \frac{x^a}{e^x} \leqq \frac{x^{\lceil a \rceil}}{e^x} < \frac{x^{\lceil a \rceil}}{\frac{x^{\lceil a\rceil+1}}{(\lceil a\rceil+1)!}} = \frac{(\lceil a\rceil+1)!}{x} < \varepsilon$$ as soon as\, $x > \max\{1, \frac{(\lceil a\rceil+1)!}{\varepsilon}\}$.\, Here, $\lceil\cdot\rceil$ \PMlinkescapetext{means} the ceiling function;\, $e^x$ has been estimated downwards by taking only one of the all positive \PMlinkescapetext{terms of the series expansion} $$e^x = 1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots$$\\ \textbf{Theorem.} The \PMlinkescapetext{growth} of the real exponential function\,\, $x\mapsto b^x$\,\, exceeds all power functions, i.e. $$\lim_{x\to\infty}\frac{x^a}{b^x} = 0$$ with $a$ and $b$ any \PMlinkescapetext{constants},\, $b > 1$. {\em Proof.}\, Since\, $\ln b > 0$,\, we obtain by using the lemma the result $$\lim_{x\to\infty}\frac{x^a}{b^x} = \lim_{x\to\infty}\left(\frac{x^{\frac{a}{\ln b}}}{e^x}\right)^{\ln b} = 0^{\ln b} = 0.$$\\ \textbf{Corollary 1.}\, $\displaystyle\lim_{x\to 0+}x\ln{x} = 0.$ {\em Proof.}\, According to the lemma we get $$0 = \lim_{u\to\infty}\frac{-u}{e^u} = \lim_{x\to 0+}\frac{-\ln{\frac{1}{x}}}{\frac{1}{x}} = \lim_{x\to 0+}x\ln{x}.$$\\ \textbf{Corollary 2.}\, $\displaystyle\lim_{x\to\infty}\frac{\ln{x}}{x} = 0.$ {\em Proof.}\, Change in the lemma\, $x$\, to\, $\ln{x}$.\\ \textbf{Corollary 3.}\, $\displaystyle\lim_{x\to\infty}x^{\frac{1}{x}} = 1.$ \, (Cf. limit of nth root of n.) {\em Proof.}\, By corollary 2, we can write:\, $\displaystyle x^{\frac{1}{x}} = e^{\frac{\ln{x}}{x}}\longrightarrow e^0 = 1$\, as\, $x\to\infty$ (see also theorem 2 in limit rules of functions).