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Proving function Converges

  1. Oct 9, 2013 #1
    1. The problem statement, all variables and given/known data
    How do I prove the function ((2^x)-1)sin(y))/(xy) converges to ln(2) in the case of x=0 and y=0..?


    2. Relevant equations



    3. The attempt at a solution
    Yeah I'm trying to figure out where to start honestly... I know i'm not suppose to post unless I have a real attempt at the solution.. I could BS something but anyone wanna point me in the right direction? Squeeze Theorem? Is there some key property I should be aware of?
     
  2. jcsd
  3. Oct 9, 2013 #2

    LCKurtz

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    Why do you think the answer is ##\ln 2##? Try some special limits to get a feel for it. What if ##y\rightarrow 0## before ##x##, for example.
     
  4. Oct 9, 2013 #3
    the answer is ln(2), the question says that, we just have to PROVE it is.
     
  5. Oct 10, 2013 #4
    Yes, the answer is indeed ln(2) I used mclaurin series to prove it.
     
  6. Oct 10, 2013 #5

    pasmith

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    Your function is a function of x only times a function of y only:
    [tex]
    \frac{(2^x - 1)\sin y}{xy} = \frac{2^x - 1}{x} \frac{\sin y}{y}
    [/tex]

    One can show that if [itex]\lim_{x \to 0} f(x)[/itex] and [itex]\lim_{y \to 0} g(y)[/itex] exist then
    [tex]
    \lim_{(x,y) \to 0} f(x)g(y) = \left(\lim_{x \to 0} f(x) \right)\left( \lim_{y \to 0} g(y) \right)
    [/tex]

    Also remember that
    [tex]
    \lim_{x \to 0} \frac{f(x) - f(0)}x = f'(0)
    [/tex]
    assuming the limit exists.
     
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