Monday, November 30, 2015

Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 26

Show that the statement limx0x3=0 is correct using the ε, δ definition of limit.

Based from the defintion,


xif 0<|xa|<δ then |f(x)L|<εxif 0<|x0|<δ then |x30|<ε



That is,x if 0<|x|<δ then |x3|<ε

Or, taking the cube root of both sides of the inequality |x3|<ε , we get...
if 0<|x|<δ then |x|<3ε

The statement suggests that we should choose δ=3ε

By proving that the assumed value of δ=3ε will fit the definition...



if 0<|x|<δ then, x|x3|<δ3=(3ε)3=ε



Thus, xif 0<|x|<δ then |x3|<εTherefore, by the definition of a limitxlimx0x3=0

No comments:

Post a Comment