growth of exponential function


Lemma.

limxxaex=0

for all values of a.

Proof.  Let ε be any positive number.  Then we get:

0<xaexxaex<xaxa+1(a+1)!=(a+1)!x<ε

as soon as  x>max{1,(a+1)!ε}.  Here, the ceiling function;  ex has been estimated downwards by taking only one of the all positive

ex=1+x1!+x22!++xnn!+

Theorem.

The of the real exponential functionDlmfDlmfMathworldPlanetmath   xbx   exceeds all power functionsDlmfDlmfPlanetmath, i.e.

limxxabx=0

with a and b any ,  b>1.

Proof.  Since  lnb>0,  we obtain by using the lemma the result

limxxabx=limx(xalnbex)lnb=0lnb=0.

Corollary 1.limx0+xlnx=0.

Proof.  According to the lemma we get

0=limu-ueu=limx0+-ln1x1x=limx0+xlnx.

Corollary 2.limxlnxx=0.

Proof.  Change in the lemma  x  to  lnx.

Corollary 3.limxx1x=1.   (Cf. limit of nth root of n.)

Proof.  By corollary 2, we can write:  x1x=elnxxe0=1  as  x (see also theorem 2 in limit rules of functions).

Title growth of exponential function
Canonical name GrowthOfExponentialFunction
Date of creation 2013-03-22 14:51:32
Last modified on 2013-03-22 14:51:32
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 18
Author pahio (2872)
Entry type Theorem
Classification msc 26A12
Classification msc 26A06
Related topic MaximalNumber
Related topic LimitRulesOfFunctions
Related topic NaturalLogarithm
Related topic AsymptoticBoundsForFactorial
Related topic MinimalAndMaximalNumber
Related topic FunctionXx
Related topic Growth
Related topic LimitsOfNaturalLogarithm
Related topic DerivativeOfLimitFunctionDivergesFromLimitOfDerivatives