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Prove continuity by first principle

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove that f(x) = x^2 is continuous at x = 2 using the ε - ∂ definition of continuity.

    2. The attempt at a solution

    Using the definition of continuity, I've reached thus far in the question:

    |x - 2||x + 2| < ε whenever |x - 2| < ∂

    3. Relevant equations

    I have no clue how to move forward from here. I know while solving this type of questions, we try to solve the first inequality so that ∂ can be written in terms of ε, but I can't seem to figure out what to do with |x + 2|.
     
  2. jcsd
  3. Oct 5, 2011 #2

    HallsofIvy

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    Okay, [itex]|x-2||x+2|< \epsilon[/itex] and so [itex]|x-2|< \epsilon/|x+ 2|[/itex]. That is almost what you want- you just need to replace that |x+2| with a constant. To do that, take some reasonable restriction on |x- 2|.

    We want to take the limit as x goes to 2 so we really are only interested, say, in |x- 2|< 1 ("1" was chosen pretty much arbitrarily). That is the same as saying that -1<x- 2< 1 and so 3< x+ 2< 5. Since 3 itself is larger than 0, that is 3< |x+2|< 5. 1/5< 1/|x+2|< 1/3 so [itex]\epsilon/5< \epsilon/|x+2|<\epsilon/3[/itex]. Now, what must your [itex]\delta[/itex] be to guarentee that "if [itex]|x- 2|< \delta[/itex], then [itex]|x- 2|< \epsilon/|x+2|[itex]"?
     
  4. Oct 5, 2011 #3
    Then [itex]\delta[/itex] = [itex]\epsilon[/itex]/5.
     
  5. Oct 5, 2011 #4
    Oh, and I totally forgot, thanks a lot for your help! :)
     
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