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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]
    |X-Xo|<[tex]\delta[/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...

    Mike
     
  2. jcsd
  3. Oct 9, 2007 #2

    HallsofIvy

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    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

    HallsofIvy

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    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]?
     
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