Limit help, confused

  1. Prove; limit as x->1 (x^3-5x+6) = 2, epsilon=0.2

    I got |x^3-5x+6-2|<0.2

    Then I don't know where to go from there. Should I add 2 to 0.2 first or subtract 2 from 6 to get x^3-5x+4 < 0.2 ?

    I'm on mobile can't use latex. Thanks
     
    Last edited: Sep 10, 2011
  2. jcsd
  3. What do you mean by "epsilon = 2"? Are you supposed to find a suitable delta?
     
  4. Whoops I meant 0.2 Yea I'm trying to find delta.
     
  5. Since you somehow want to use the fact that |x-1| < [itex]\delta[/itex], you should be trying to divide this polynomial by x-1.
    Then you'll have |x3-5x+4| = |x-1| |P(x) | < [itex]\delta[/itex] |P(x)|, where P(x) is some other polynomial of order 2, which you could also bound...

    Try that. :)
     
    Last edited: Sep 10, 2011
  6. I'm sorry but I still do not understand. Couldn't we just solve for epsilon then replace it for delta? Thanks
     
  7. I don't exactly understand what you mean.

    You wrote |x^3-5x+6-2|<0.2
    Defining f(x)= x3 - 5x + 6
    But it's not that you need to solve an inequality. What you need to do, formally, is to find a [itex]\delta[/itex], so that for every x that holds |x-1| < [itex]\delta[/itex],
    |f(x) - 2|| < 0.2.

    So, like I said, you have to somehow use the fact that |x-1| < [itex]\delta[/itex].
    You start off by writing |x3 - 5x + 6 - 2|, and you start looking for ways to somehow express it with, apart from other things, [itex]\delta[/itex]. You are looking to see how small this [itex]\delta[/itex] needs to be, so that |x3 - 5x + 6 - 2| is small enough.

    You start off in the way I wrote. Otherwise - I don't understand your question.
     
  8. Well it asks me use the graph to find a number Delta.

    If |x-1| < d then |(x^3-5x+6)-2|<0.2

    Then it tells me find Delta that corresponds to epsilon =0.2

    Oh ok, Im starting to see it.
     
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