growth of exponential function

Lemma.

$\displaystyle \lim_{x\to\infty}\frac{x^a}{e^x} = 0$

for all constant values of $ a$ .

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

$\displaystyle 0 < \frac{x^a}{e^x} \leqq \frac{x^{\lceil a \rceil}}{e^x} <
\frac...
...a\rceil+1}}{(\lceil a\rceil+1)!}}
= \frac{(\lceil a\rceil+1)!}{x} < \varepsilon$

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

$\displaystyle e^x = 1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots$


theorem. The growth of the real exponential function $ x\mapsto b^x$ exceeds all power functions, i.e.

$\displaystyle \lim_{x\to\infty}\frac{x^a}{b^x} = 0$

with $ a$ and $ b$ any constants, $ b > 1$ .

Proof. Since $ \ln b > 0$ , we obtain by using the lemma the result

$\displaystyle \lim_{x\to\infty}\frac{x^a}{b^x} =
\lim_{x\to\infty}\left(\frac{x^{\frac{a}{\ln b}}}{e^x}\right)^{\ln b} = 0^{\ln b} = 0.$


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

Proof. According to the lemma we get

$\displaystyle 0 = \lim_{u\to\infty}\frac{-u}{e^u} =
\lim_{x\to 0+}\frac{-\ln{\frac{1}{x}}}{\frac{1}{x}} = \lim_{x\to 0+}x\ln{x}.$


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

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

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

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).



Contributors to this entry (in most recent order):

As of this snapshot date, this entry was owned by pahio.