continuity of natural power


Theorem.

Let n be arbitrary positive integer.  The power functionxxn  from to (or to ) is continuous at each point x0.

Proof.  Let ε be any positive number.  Denote  x0+h=x  and  xn-x0n=Δ.  Then identically

Δ=(x-x0)(xn-1+xn-2x0++x0n-1).

Taking the absolute valueMathworldPlanetmathPlanetmathPlanetmath and using the triangle inequalityMathworldMathworldPlanetmathPlanetmath give

|Δ|=|h||xn-1+xn-2x0++x0n-1||h|(|xn-1|+|xn-2x0|++|x0n-1|).

But since  |x|=|x0+h||x0|+|h|  and also  |x0||x0|+|h|,  so each summand in the parentheses is at most equal to  (|x0|+|h|)n-1,  and since there are n summands, the sum is at most equal to n(|x0|+|h|)n-1.  Thus we get

|Δ|n|h|(|x0|+|h|)n-1.

We may choose  |h|<1;  this implies

|Δ|n|h|(|x0|+1)n-1.

The right hand side of this inequalityMathworldPlanetmath is less than ε as soon as we still require

|h|<εn(|x0|+1)n-1.

This means that the power function  xxn is continuous at the point x0.

Note.  Another way to prove the theorem is to use induction on n and the rule 2 in limit rules of functions.

Title continuity of natural power
Canonical name ContinuityOfNaturalPower
Date of creation 2013-03-22 15:39:25
Last modified on 2013-03-22 15:39:25
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 7
Author pahio (2872)
Entry type Theorem
Classification msc 26C05
Classification msc 26A15
Related topic Exponentiation2