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Finding a limit

  1. Jul 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the lim x-->0 x^2sin(e/x).


    2. Relevant equations



    3. The attempt at a solution
    Using the limit laws I can say that
    lim x-->0 x^2sin(e/x) = lim x-->0 x^2 * lim x-->0 sin(e/x)
    lim x-->0 x^2=0; but lim x-->0 sin(e/x) however, is undefined...right?

    So, the limit doesn't exist?
     
  2. jcsd
  3. Jul 10, 2008 #2

    quasar987

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    The theorem is... IF lim f and lim g exist, then lim(f+g) exists and equals limf+limg. Here, lim sin(e/x) doesn not exist, so you can't use this theorem.

    But try to use the facts that |sin(y)|<=1 and the fact that lim |f(x)|=0 <==> lim f(x)=0.
     
  4. Jul 10, 2008 #3

    Tom Mattson

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    No, you can't say that. You have to read the limit law carefully. It only allows you to split the limits up like that when both limits do exist.

    Instead try the following change of variables. Let [itex]u=\frac{1}{x}[/itex]. Then [itex]x\rightarrow 0 \Rightarrow u\rightarrow\infty[/itex].
     
  5. Jul 10, 2008 #4
    Could you clarify this a bit more for me, please? Should I be subsituting [itex]u=\frac{1}{x}[/itex] for (e/x)? And essentially by taking the sin(infinity) the limit would be -1? Am I following this correctly?

    Forgive me if I'm a little slow with this. I'm a newbie to calculus, this is the first term I've ever taken and it's an online accelerated course. So, while I enjoy math and usually get it pretty quickly, my extreme overload of information has got my brain sort of going in circles.

    And, thank you! I really appreciate the help. =)
     
  6. Jul 10, 2008 #5

    nicksauce

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    He means that you should consider
    [tex]\lim_{u\rightarrow\infty}\frac{1}{u^2}sin(ue)[/tex]

    You can then apply the squeeze theorem to this limit.


    "And essentially by taking the sin(infinity) the limit would be -1?"
    Not sure where you get this. sin(infinity) is not well-defined in any sense.
     
  7. Jul 11, 2008 #6

    HallsofIvy

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    You might also use the fact that
    [itex]\lim_{x\rightarrow 0}\frac{sin(x)}{x}= 1[/itex]
     
  8. Jul 11, 2008 #7
    Halls: can't he use squeeze theorem directly without the substitution, or am I missing something?
     
  9. Jul 11, 2008 #8

    HallsofIvy

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    Yes, you could. I was just suggesting another possibility.
     
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