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Precise Definition of a Limit

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Using the epsilon and delta definition, prove that:
    [tex]\mathop {\lim }\limits_{x \to - 3} \sqrt {{x^2} + 16} = 5[/tex]

    2. Relevant equations



    3. The attempt at a solution

    Given epsilon > 0. Choose [tex]\delta {\rm{ = min}}\left\{ {{\rm{1,}}\frac{{\left( {5 + \sqrt {20} } \right)\varepsilon }}{7}} \right\}[/tex], then:
    [tex]0 < \left| {x + 3} \right| < \delta \Rightarrow \left| {\sqrt {{x^2} + 16} - 5} \right| = \frac{{\left| {x + 3} \right|.\left| {x - 3} \right|}}{{5 + \sqrt {{x^2} + 16} }}[/tex]
    Moreover, [tex]\left| {x + 3} \right| < 1 \Rightarrow \left| {x - 3} \right| < 7[/tex] and [tex]5 + \sqrt {{x^2} + 16} > 5 + \sqrt {20} [/tex]
    Hence, [tex]\frac{{\left| {x + 3} \right|.\left| {x - 3} \right|}}{{5 + \sqrt {{x^2} + 16} }} < \frac{{\frac{{\left( {5 + \sqrt {20} } \right)\varepsilon }}{7}.7}}{{5 + \sqrt {20} }} = \varepsilon .[/tex]This completes the proof.

    Please correct if there's anything wrong with it. Thanks!
     
    Last edited: Aug 30, 2012
  2. jcsd
  3. Aug 30, 2012 #2
    Looks good.
     
  4. Aug 30, 2012 #3

    Bacle2

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    Small nitpick on calculations:

    in part 3, after "then" , at the end, the factors should be |x-5||x+5|.
     
  5. Aug 30, 2012 #4
    How did you get that?
     
  6. Aug 30, 2012 #5

    SammyS

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    You may want to recheck that.

    [itex](\sqrt{x^2+16}\ )^2 - 5^2=x^2+16-25=x^2-9=\dots[/itex]
     
  7. Aug 30, 2012 #6

    Bacle2

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    Never mind, my bad. Computation mistake from jumping-in too quickly.
     
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