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Identify the value of δ( accur. to 2 dec. places) that corresponds to ε=0.01?

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Identify the value of δ( accur. to 2 dec. places) that corresponds to ε=0.01, given lim x->-1 (x^2+3)=4, according to the definition of limits.


    2. Relevant equations

    |f(x)- L| < ε

    0<|x-a| < δ

    3. The attempt at a solution

    |(x^2+3)- 4| <0.01

    |x^2-1| <0.01

    |x-1||x+1| <0.01

    This where I have no idea what to with |x-1|. Do I divide 0.01 by |x-1|? Any hints, please?

    Thanks.
     
  2. jcsd
  3. May 13, 2012 #2

    HallsofIvy

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    Staff Emeritus
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    You don't want to divide by |x- 1|, you want to divide by a number which means that you need to replace |x- 1| by a number that keeps the inequality true.

    You only need to prove this for x "close" to -1 so start by assuming that |x-(-1)|= |x+ 1|< 1 (1 is chosen just as a convenient value- any positive number would work). That is, that -1< x+1< 1. Subtracting 2 from each part, -3< x- 1< -1 which is the same as saying that |x- 1|> 1. We want ">" here because now we can say that if we replace |x- 1| by 1 in |x- 1||x+ 1| we have replace it by something larger: |x-1||x+1|< |x+ 1|. If we take |x+1|< .001 then it is certainly true that |x-1||x+1|< |x+1|< .001.
     
  4. May 16, 2012 #3
    Hi, HallsOfIvy. I appreciate your answer. Just never got around to answering to this thread. Wanted to acknowledge I read your answer and probably will have some questions once I finally have time to look at it a bit closer. Thank You, HallsOfIvy.
     
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