elementary proof of growth of exponential function

Proposition 1.

If x is a non-negative real number and n is a non-negative integer, then (1+x)n1+nx.


When n=0, we have (1+x)0=11+00. If, for some natural numberMathworldPlanetmath n, it is the case that (1+x)n1+nx then, multiplying both sides of the inequalityMathworldPlanetmath by (1+x), we have


By inductionMathworldPlanetmath, (1+x)n1+nx for every natural number n. ∎

Proposition 2.

If b is a real number such that b>1 and n and k are non-negative integers, we have bn>(b-1bk)knk.


Let x=b-1. Write n=mk-r where m and r are non-negative integers and r<k.

By the preceding propositionPlanetmathPlanetmathPlanetmath, (1+x)m>mx. Raising both sides of this inequality to the kth power, we have (1+x)mk>(mx)k. Since r<k, we also have (1+x)-r>(1+x)-k; multiplying both sides by this inequality and collecting terms,


Multiplying the right-hand side by kk/kk and rearranging,


Since mkmk-r, we also have


Recalling that mk-r=n and 1+x=b, we conclude that


Proposition 3.

If a, b, and x are real numbers such that a0, b>1 and x>0, then


Let k and n be integers such that ak<a+1 and xnx+1. Since x+1>n, we have bx+1>bn. By the preceeding proposition, we have


Since k<a+1, we have 1/kk>1/(a+1)k, so


Since ka0, we have


Summarrizing our progress so far,


Dividing both sides by b and simplifying,


Proposition 4.

If a and b are real numbers and b>1, then


Substituting a+1 for a


Dividing by x and rearranging,


Since limx0=0 and limx1x=0, we also have limxxabx=0 by the squeeze rule.

Title elementary proof of growth of exponential function
Canonical name ElementaryProofOfGrowthOfExponentialFunction
Date of creation 2014-03-10 17:57:26
Last modified on 2014-03-10 17:57:26
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 28
Author rspuzio (6075)
Entry type Definition
Classification msc 26A12
Classification msc 26A06