Limit of two-variable function

[twoflower]

Hi all,

suppose I want to get this:

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

Here's how I approached:

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

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

<br /> \lim_{t \rightarrow 0+} t\ \log t = 0<br />

Thank you for help.

 

[VietDao29]

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