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Does anybody know how to prove that in $D=\{(x,y)\in\mathbb{R}^2:x>0\wedge y>0\}$ the following is true: $$ \lim\limits_{(x,y)\to(0,0)}x\cdot y\cdot\ln{(x\cdot y)}=0 $$ I have to find a $\delta$ so that if $\|(x,y)\|=\sqrt{x^2+y^2}<\delta$, that $|x\cdot y\cdot\ln{(x\cdot y)}|<\epsilon$ follows. But I don't know what to do, because the $\ln$ goes to minus infinity.

Can anybody solve this? Thank you!

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Can't you use L'Hospital's rule? –  draks ... Jun 20 '12 at 8:54
    
Hint: $t \ln t \to 0$ as $t \to 0^+$. –  Hans Lundmark Jun 20 '12 at 8:59
    
But how do I use the L'Hopital's rule in R^2? I thought it was only for normal limits, not for multivariable, because where to i have to differentiate to then? X or Y? –  Carucel Jun 20 '12 at 9:01
    
@Carucel: Do you have to find $\delta$ which $\delta^{\delta}=e^{\epsilon}$? –  B. S. Jun 20 '12 at 9:03

1 Answer 1

up vote 1 down vote accepted

The key is to treat $xy$ as one variable. Let $z = xy$. Hence, $$\lim_{(x,y) \to (0^+,0^+)} xy \ln(xy) = \underbrace{\lim_{z \to 0^+} z \ln(z) = -\lim_{t \to \infty} t e^{-t}}_{z = e^{-t}}$$ Note that $e^t \geq \dfrac{t^2}{2}$, for $t \geq 0$. $\left(\text{$\dfrac{t^2}2$ is a term in Taylor series of $e^t$ and all the other terms are non-negative for $t >0$} \right).$

Hence, $$t e^{-t} = \dfrac{t}{e^t} \leq \dfrac{t}{t^2/2} = \dfrac2t$$ Hence, $$0 \leq \lim_{t \to \infty} t e^{-t} \leq \lim_{t \to \infty} \dfrac2t = 0$$

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It is the substitution xy=z which makes this limit a simple one. Ok, thank you very much! –  Carucel Jun 20 '12 at 9:03

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