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Another limit problem.

  1. Oct 2, 2004 #1
    lim_x->0 (x + (1/x) ) sin(x)

    As x goes towards 0, wouldn't 1/x go towards inf?
    Thus
    (x + (1/x)) should go towards inf as x goes towards 0?

    sin(x) will go towards 0 as x goes towards 0, since sin(0) is 0.

    Wouldn't that leave inf * 0 = undefined (or 0)?
     
  2. jcsd
  3. Oct 2, 2004 #2

    arildno

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    Note that your expression is the sum of two other expressions:
    [tex](x+\frac{1}{x})\sin(x)=x\sin(x)+\frac{\sin(x)}{x}[/tex]
    1. If it can be shown that each of these expressions (on the right-hand side) has a limit as x->0, what's then true about the limit of their sum (i.e, your original expression)?
     
  4. Oct 2, 2004 #3
    lim_x->0 f(x) = x*sinx = 0
    lim_x->0 g(x) = sinx/x = 1

    lim_x->0 f(x) + g(x) = 0 + 1

    lim_x->0 (x + 1/x))sinx = 1

    Hm, that was a lot easier than I thought, thanks a lot :)

    Edit: How do you do latex?
    test:
    [itex]\pi[/itex]

    nevermind :)
     
  5. Oct 2, 2004 #4
    ah, just l'hopital the second part :-)
     
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