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Limit proof

  1. Jan 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove [tex]lim_{x->4}\frac{x-4}{\sqrt{x}-2}=4[/tex]

    2. Relevant equations
    Epsilon\delta definition

    3. The attempt at a solution

    I can see that a direct evaluation at 4 leads to an indeterminate form, so:

    [tex]\frac{x-4}{\sqrt{x}-2}*\frac{\sqrt{x}+2}{\sqrt{x}+2} \; \mbox{simplifies to} \; \sqrt{x}+2 \; \mbox{when} \; x \ne 4[/tex]

    Via epsilon\delta definition,
    [tex]|\sqrt{x}+2-4|<\epsilon \; \mbox{whenever} \; |x-4|<\delta[/tex]

    Expanding the right side,

    [tex]-\delta<x-4<\delta[/tex]
    [tex]-\delta+4<x<\delta+4[/tex]
    [tex]\sqrt{-\delta+4}<\sqrt{x}<\sqrt{\delta+4} \; \mbox{when} \; \delta \leq 4[/tex]
    [tex]\sqrt{-\delta+4}-2<\sqrt{x}-2<\sqrt{\delta+4}-2[/tex]

    Thus, [tex]\delta=min\{\sqrt{-\delta+4}-2,4,\sqrt{\delta+4}-2\}[/tex]

    Given this definition of delta, it is elementary to work backwards towards [tex]|\sqrt{x}+2|<\delta[/tex]

    Have I made any errors in my proof?
     
  2. jcsd
  3. Jan 19, 2015 #2

    perplexabot

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    Gold Member

    Hey cuibba. I can't comment on your technique, but you could have used L'hopital's rule. That way I think is easier.
     
  4. Jan 19, 2015 #3

    pasmith

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    Homework Helper

    What is [itex]\delta[/itex] when [itex]\epsilon = 10[/itex]? What about when [itex]\epsilon = 10^{-5}[/itex]?

    Your task is to prove:

    For every [itex]\epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that if [itex]0 < |x - 4| < \delta[/itex] then [itex]|\frac{x - 4}{\sqrt{x} - 2} - 4| < \epsilon[/itex].

    You need to relate your [itex]\delta[/itex] to the [itex]\epsilon[/itex].
     
  5. Jan 19, 2015 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    He could have used l'Hospital's rule (yes, it really is spelled Hospital) but his way is easier. Alternatively, he could have set ##\sqrt{x} = y## and then evaluated
    [tex] \lim_{y \to 2} \frac{y^2 - 4}{y-2},[/tex]
    because in this form l'Hospital's rule is much nicer to use.
     
  6. Jan 19, 2015 #5
    Ah, I wish I was allowed to use that technique, but our prof. is adamant on teaching delta ep. definitions first.

    Oops, I mistyped. I meant to say "Given this definition of delta, it is elementary to work backwards towards [tex]
    |\sqrt{x}+2\mathbf{-4}|<\boldsymbol{\epsilon}
    [/tex] My main question is whether or not it is correct to say that, for any epsilon, delta is [tex]
    min\{\sqrt{-\delta+4}-2,4,\sqrt{\delta+4}-2\}[/tex]


    Ah, I didn't think to do a u-sub. I'll definitely try that next time!
     
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