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Prove by definition the statement lim x->4 √x = 2

  1. Apr 2, 2008 #1
    Prove by definition the statement

    lim x->4 √x = 2

    This is what I've done:
    Given ε>0

    |(√x)-2| = |√x-2|< ε if |x-4|< ε^2

    I choose δ = ε^2

    0<|x-4|< ε^2
    => |x-2|< √ε^2 = ε

    Hence lim x->4 √x = 2

    So, what do you think? Did I do the process correctly?
     
    Last edited: Apr 2, 2008
  2. jcsd
  3. Apr 2, 2008 #2
    Well what you actually need to prove is that

    [tex]\forall\epsilon>0,\exists\delta(\epsilon)>0[/tex] such that whenever

    [tex]0<|x-4|<\delta=>|\sqrt{x}-2|<\epsilon \ \ ??[/tex]

    [tex]\lim_{x\rightarrow 4}\sqrt{x}=2[/tex]

    so

    [tex]|(\sqrt{x}-2)\frac{\sqrt{x}+2}{\sqrt{x}+2}|=|\frac{x-4}{\sqrt{x}+2}|<\frac{\delta}{\sqrt{x}+2}<\frac{\delta}{2}=\epsilon=>\delta=2\epsilon[/tex]

    Now since x-->4, it is safe to assume that

    0<x<5 , now lets take the square root of both sides

    [tex]0<\sqrt{x}<\sqrt{5} /+2=>2<\sqrt{x}+2<\sqrt{5}+2=>\frac{1}{\sqrt{x}+2}<\frac{1}{2}[/tex]

    Now to be absolutely sure that this holds we choose [tex]\delta=min(2\epsilon, 1)[/tex]

    This is all we needed. Do you understand how it goes now?
     
    Last edited: Apr 2, 2008
  4. Apr 2, 2008 #3
    Yes, I understand it now. Thank you!

    My other question is; consider: lim x->0 x^2sin(1/x)

    What do you think of the limit (i.e. it exists or it's infinite) ?
    How can we prove that this limit is/isn't infinite? What's the working?

    Thanks.
     
  5. Apr 2, 2008 #4

    HallsofIvy

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    How does that follow? Surely you didn't take the square root of both sides! This is the crucial part of your proof and it is not at all obvious. It isn't even true without other conditions: if x= 5, [itex]\epsilon= 2[/itex], then 0<|x-4|= 1< [itex]\epsilon^2[/itex]= 4 but |x- 2|= 3 is NOT less than [itex]\epsilon[/itex]= 2.

     
    Last edited: Apr 2, 2008
  6. Apr 2, 2008 #5

    Well, what are your thoughts on it? Show what you did so far, what have you tried?
     
  7. Apr 2, 2008 #6

    HallsofIvy

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    If the sin(1/x) is bothering you, let y= 1/x and replace all xs with y.
     
  8. Apr 3, 2008 #7
    Yes, I have also tried another question, I don't know if I'm right but I'll try it:
    lim x->∞ (1+e^x)/e^x

    I have sketched the graph and I think the limit is 1.

    Now to prove the limit:

    Let ε>0 be given.

    We choose K = ln(1/ε)>0
    so that;
    x>k = ln(1/ε) => 1/ε<e^x => e^-x <ε
    |((1+e^x)/e^x)|-1 = e^-x< ε

    Please correct me if I'm wrong. I'd appreciate that.
     
  9. Apr 3, 2008 #8

    HallsofIvy

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    Perhaps it would be simpler as (1+ ex)/ex= e-x+ 1.
     
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