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Limit of two-variable function

  1. Oct 14, 2005 #1
    Hi all,

    suppose I want to get this:

    \lim_{[x,y] \rightarrow [0,0]} (x^2+y^2)^{xy}

    Here's how I approached:

    \lim_{[x,y] \rightarrow [0,0]} (x^2+y^2)^{xy} = \lim_{[x,y] \rightarrow [0,0]} \exp^{xy \log (x^2+y^2)}

    \lim_{[x,y] \rightarrow [0,0]} xy \log (x^2 + y^2) = (x^2 + y^2) \log (x^2 + y^2) \frac{xy}{x^2 + y^2} \rightarrow 0

    Because the last fraction is bounded and the part before it goes to 0 (I hope).

    But that's the problem, I don't know how to prove

    \lim_{t \rightarrow 0+} t\ \log t = 0

    Thank you for help.
  2. jcsd
  3. Oct 14, 2005 #2


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    Homework Helper

    You can use L'Hopital rule to prove that. Try rearrange the equation to:
    [tex]\lim_{t \rightarrow 0 ^ +} t \log{t} = \lim_{t \rightarrow 0 ^ +} \frac{\log{t}}{\frac{1}{t}}[/tex]. Now it's in form [tex]\frac{\infty}{\infty}[/tex]. Can you go from here?
    Viet Dao,
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