1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Real Analysis: Proving a fuction is continuous

  1. Oct 9, 2007 #1
    I Apologize in advance for the amount of questions i plan on asking you guys this year. Real Analysis is my first upper division math class and i have not trained my mind to think abstractly enough yet.

    1. The problem statement, all variables and given/known data

    i) Show that f(x) = x^3 is continuous on R by using (Epsilon)−(Delta) property

    3. The attempt at a solution

    Ok so there is basically one part that i get stuck at, but here is what i have so far

    Suppose that limit Xn= Xo

    So we have lim f(Xn)=lim[Xn^3]= Xo^3 =f(Xo)
    hence f is continuous for each Xo in R

    |f(X)-f(Xo)|=|X^3 - Xo^3|= |(X - Xo)(X^2 + XXo + Xo^2)| <[tex]\epsilon[/tex]

    Now here is where i am stuck. How exactly do i find epsilon? Have I even done this proof correctly so far? Also after I find epsilon, what exactly do i do...

    Please Reply Over...

  2. jcsd
  3. Oct 9, 2007 #2


    User Avatar
    Science Advisor

    So you have [itex]|x- x_0||x^2+ x_0x+ x_0^2|< \epsilon[/itex] and the problem is that second absolute value.

    Choose some convenient distance from x0. For example, let's suppose that
    [itex]|x- x_0|< 1[/itex]. That is the same as saying [itex]x_0-1< x< x_0+ 1[/itex]. Use that to find an upper bound on [itex]x^2+ x_0x+ x_0^2[/itex], say M, so you have
    [itex]|x-x_0||x^2+x_0x+ x_0^2|< M|x-x_0|[/itex]. That will be less than [itex]\epsilon[/itex] as long as [itex]M|x-x_0|< \epsilon[/itex] which says [itex]|x-x_0|< \frac{\epsilon}{M}[/itex]. All you have to do is decide how large M has to be.
  4. Oct 10, 2007 #3
    Ok the highlighted statment is giving me some trouble... wouldnt M just be equal to [itex]|x^2 + x_0x+ x_0^2|[/itex] ?
  5. Oct 11, 2007 #4


    User Avatar
    Science Advisor

    No, because M has to be a number, not a function of x. If x is between [itex]x_0-1[/itex] and [itex]x_0+ 1[/itex] what is the largest possible value of [itex]x^2+ x_0x+ x_0^2[/itex]?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook