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Basic limit question ( f(x) = x^2 )

  1. Mar 22, 2012 #1
    f(x) = x^2 and we want to show that as x -> a, f(x) -> a^2.

    So, we must ensure that |x^2 - a^2| < eps.

    The book (Calculus by Spivak) starts of by bounding |x - a| < 1 which leads us to
    |x + a| < 2|a| + 1.

    This is the part where I get lost:
    I can understand that we require |x - a| < eps/ (2|a| + 1),
    but
    why do we choose |x - a| < min(1, eps/ (2|a| + 1)) ?

    And what happens if min(1, eps/ (2|a| + 1)) = 1 ?

    Thank you very much for your help in advance! :)
     
  2. jcsd
  3. Mar 22, 2012 #2

    tiny-tim

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    hi ltjrpliskin! :smile:

    (try using the X2 button just above the Reply box :wink:)

    |x2 - a2| = |x - a||x + a|,

    so we need |x + a| to be bounded

    any bound would do, but it seems most sensible to use 1 :wink:
     
  4. Mar 22, 2012 #3
    Okay, but can you help me understand why we chose the minimum of 1 and eps/ (2|a| + 1)? :)
     
  5. Mar 22, 2012 #4

    tiny-tim

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    we want |x - a||x + a| to be small …

    |x - a| < 1 keeps |x + a| small, in fact < (2|a| + 1)

    |x - a| < ε/ (2|a| + 1), keeps |x - a| small,

    and both together keep the whole thing < ε :wink:

    so we need both conditions … it won't work if |x - a| can be unbounded
     
  6. Mar 22, 2012 #5
    Ahh! Okay, thank you very much :)
    That was much simpler than expected haha :)
     
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