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Homework Help: Limit problem. Chapter :Precise definition

  1. Mar 25, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the limit

    lim x[tex]\rightarrow[/tex]1 (4+x-3(x^3) = 2

    find the values of [tex]\delta[/tex] that correspond to [tex]\epsilon[/tex]=1 and

    [tex]\epsilon[/tex] = 0.1


    2. Relevant equations



    3. The attempt at a solution
    As much as I know I should do like this.

    |(4 + x - 3(x^3)) - 2| < 1 /////// |x-1| < [tex]\delta[/tex]

    |-3(x^3) + x + 2| < 1 so what so on?


    I have no idea.
    Thank you in advance.]
    Please if you can give me aa short explan with the steps.
    Thanks thanks.
     
  2. jcsd
  3. Mar 25, 2008 #2
    well this does not look that bad at all: The idea is quite simple indeed. Here is what you need to do

    [tex]|4+x-3x^{3}-2|<\epsilon=1[/tex] frome here using the properties of the absolute value we have:

    [tex]-1<2+x-3x^{3}<1[/tex] now let this expression be equal to -1, and 1 ,and solve for x. ALso after that use the fact that

    [tex]|x-1|<\delta=>-\delta<x-1<\delta =>1-\delta<x<1+\delta[/tex]. then use these two expression to come up with the value of delta.

    Do the exact same steps for epsilon=0.1

    This is almost the whole thing, all i have left out are calculations, simply algebra steps. I think you will be fine from here.

    Can you take it from here???
     
    Last edited: Mar 26, 2008
  4. Mar 25, 2008 #3
    Not really.
     
  5. Mar 26, 2008 #4
    An ideal way would be to manage to solve this eq. [tex]2+x-3x^{3}=1[/tex] but since there seem to be no obvious solutions, than doing it is going to be a pain. But i think we can proceede this way also, i am not that sure though:
    [tex]-|3x^{3}|=-|x||3x^{2}|\leq-|x||3x^{2}-1|\leq2-|x||3x^{2}-1|\leq|2-x(3x^{2}-1)|<\epsilon=1[/tex], so [tex]|3x^{3}|>1=>-1>3x^{3}>1=>-\frac{1}{3}>x^{3}>\frac{1}{3}[/tex] the corresponging equation

    [tex]x^{3}=\pm \frac{1}{3}[/tex] will have three roots, one real, and two complex. so lets take just real ones from here we get

    [tex]-\frac{1}{\sqrt[3]{3}}>x>\frac{1}{\sqrt[3]{3}}[/tex].

    Now using the fact that: [tex]1-\delta<x<1+\delta[/tex] we get: [tex]1-\delta=\frac{1}{\sqrt[3]{3}} \ \ \\ \ and \ \ \\ \ \ 1+\delta=-\frac{1}{\sqrt[3]{3}}[/tex]

    Now from here you will be able to find delta, and chose the minimum. That is whicheverone is smaller and positive. But as much as i can see there will be only one such.

    I hope this works. I am not sure that what i did is entirely correct, because that polynomial of degree 3 complicated the whole thing. Anyhow i think that this is the idea behind these types of problems.
     
  6. Mar 26, 2008 #5

    HallsofIvy

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    You know that [itex]\lim_{x\rightarrow 1} 4+ x- x^3= 2[/itex] because you know that polynomials are continuous and putting x= 1 makes [itex]4+ x- x^3[/itex] equal to 2. That in turn tells you that x= 1 makes [itex]4+ x- x^3 -2[/itex] equal to 0. And that tells you that x-1 is a factor of [itex]2+ x- x^3[/itex]. Then, by dividing, say, you can determine that [itex]2+ x- x^3= (-1)(x- 1)(3x^2+ 3x+ 2)[/itex]. If x is close to 1, we can certainly say that x is between 0 and 2: 0< x< 2 so 0< x2< 4, 0< 3x2< 12, 0< 3x< 6 and, finally, 2< 3x2+ 3x+ 2< 12+ 6+ 2= 20 and so 0< |3x2+ 3x+ 2|< 20. That is, |2+ x- x^3|= |x-1||3x^2+ 3x+ 2|< 20|x-1|.
    You can make that less than [itex]\epsilon[/itex] by making |x-1| less than [itex]\epsilon/20[/itex] (or 1 whichever is smaller).
     
  7. Mar 26, 2008 #6
    Well this would be great, only if the OP's polynomial wasn't slightly different from yours here. Your work would be valid if the question at issue actually would be [tex]\lim_{x\rightarrow 1} 4+ x- x^3= 2[/tex] while it actually is [tex]\lim_{x\rightarrow 1} 4+ x- 3x^3= 2[/tex]

    [tex]4+x-3x^{3}[/tex] It is this 3 in front of the cubic that makes things not look that nice.
     
    Last edited: Mar 26, 2008
  8. Mar 26, 2008 #7

    HallsofIvy

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    Science Advisor

    Oh blast! My eyes are going.

    Actually, everything I said is correct for [itex]4+ x- 3x^3[/itex] because that was what I was working with, even though I wrote [itex]4+ x- x^3[/itex] (which obviously does NOT have limit 2 as x goes to 1).
    [itex]4+ x- 3x^3= (-1)(x-1)(3x^2+ 3x+ 1)[/itex] and everything I said applies to that.
     
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