Limit of two-variable function

  • Thread starter twoflower
  • Start date
  • #1
368
0
Hi all,

suppose I want to get this:

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

Here's how I approached:

[tex]
\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
[/tex]

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

[tex]
\lim_{t \rightarrow 0+} t\ \log t = 0
[/tex]

Thank you for help.
 

Answers and Replies

  • #2
VietDao29
Homework Helper
1,423
3
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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