PlanetPhysics/Growth of Exponential Function

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

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, $$\lceil\cdot\rceil$$ 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. $$\lim_{x\to\infty}\frac{x^a}{b^x} = 0$$ with $$a$$ and $$b$$ any constants,\, $$b > 1$$.

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

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

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

Corollary 2. \, $$\lim_{x\to\infty}\frac{\ln{x}}{x} = 0.$$

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

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

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