proof of growth of exponential function


In this proof, we first restrict to when x and a are integers and only later lift this restricton.

Let a>0 be an integer, let b>1 be real, and let x be an integer.

Consider the following inequality

(1+1x)a≤1+ax⁢(1+1x)a-1

If x≥2, then we have

(1+1x)a≤1+ax⁢(32)a-1.

Define X to be the greater of 2 and ⌈a⁢(3/2)a-1/(1-b)⌉; when x>X, we have

(1+1x)a≤b.

Rewrite xa/bx as follows when x>X:

xabx=XabX⁢∏n=Xx(1+1n)a⁢1b

By the inequality established above, each term in the productMathworldPlanetmathPlanetmath will be bounded by 1/b, hence

xabx≤XabX⁢1(b)x-X

Since b>1, it is also the case that b>1, hence we have the inequality

(b)n≥1+n⁢(b-1)

Combining the last two inequalities yields the following:

xabx≤XabX≤11+(x-X)⁢(b-1)

From this, it follows that limx→∞⁡xa/bx=0 when a and x are integers.

Now we lift the restrictionPlanetmathPlanetmathPlanetmath that a be an integer. Since the power function is increasing, xa/bx≤x⌈a⌉/bx, so we have limx→∞⁡xa/bx=0 for real values of a as well.

To lift the restriction on x, let us write x=x1+x2 where x1 is an integer and 0≤x2<1. Then we have

xabx=x1abx1⁢(x1+x2x1)a⁢b-x2

If x>2, then (x1+x2)/x2<1.5. Since x2≥0,b-x2≤1. Hence, for all real x>2, we have

xabx≤1.5a⁢x1abx1

From this inequality, it follows that limx→∞⁡xa/bx=0 for real values of x as well.

Title proof of growth of exponential function
Canonical name ProofOfGrowthOfExponentialFunction
Date of creation 2013-03-22 15:48:36
Last modified on 2013-03-22 15:48:36
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 11
Author rspuzio (6075)
Entry type Proof
Classification msc 32A05