PlanetPhysics/Growth of Exponential Function

Lemma. $\underset{x\to \mathrm{\infty }}{\lim }\frac{x^a}{e^x}=0$ for all constant values of $a$.

Proof. \, Let $\epsilon$ be any positive number.\, Then we get:

$0<\frac{x^a}{e^x}\leqq \frac{x^{\lceil a\rceil }}{e^x}<\frac{x^{\lceil a\rceil }}{\frac{x^{\lceil a\rceil +1}}{(\lceil a\rceil +1)!}}=\frac{(\lceil a\rceil +1)!}{x}<\epsilon$ as soon as\, $x>\max \{1,\frac{(\lceil a\rceil +1)!}{\epsilon }\}$.\, Here, $\lceil \cdot \rceil$ means the ceiling function;\, $e^x$ has been estimated downwards by taking only one of the all positive terms of the series expansion $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots +\frac{x^n}{n!}+\cdots$\\

\htmladdnormallink{theorem {http://planetphysics.us/encyclopedia/Formula.html}.} The growth of the real exponential function\,\, $x\mapsto b^x$\,\, exceeds all power functions, i.e. $\underset{x\to \mathrm{\infty }}{\lim }\frac{x^a}{b^x}=0$ with $a$ and $b$ any constants,\, $b>1$.

Proof. \, Since\, $\ln b>0$,\, we obtain by using the lemma the result $\underset{x\to \mathrm{\infty }}{\lim }\frac{x^a}{b^x}=\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{x^{\frac{a}{\ln b}}}{e^x}\right)}^{\ln b}=0^{\ln b}=0.$\\

Corollary 1. \, $\underset{x\to 0+}{\lim }x\ln x=0.$

Proof. \, According to the lemma we get $0=\underset{u\to \mathrm{\infty }}{\lim }\frac{-u}{e^u}=\underset{x\to 0+}{\lim }\frac{-\ln \frac{1}{x}}{\frac{1}{x}}=\underset{x\to 0+}{\lim }x\ln x.$\\

Corollary 2. \, $\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln x}{x}=0.$

Proof. \, Change in the lemma\, $x$\, to\, $\ln x$.\\

Corollary 3. \, $\underset{x\to \mathrm{\infty }}{\lim }{x}^{\frac{1}{x}}=1.$ \, (Cf. limit of nth root of n.)

Proof. \, By corollary 2, we can write:\, $x^{\frac{1}{x}}=e^{\frac{\ln x}{x}}⟶e^0=1$\, as\, $x\to \mathrm{\infty }$ (see also theorem 2 in limit rules of [[../Bijective/|functions]]).