# Limit of two-variable function

1. Oct 14, 2005

### twoflower

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. Oct 14, 2005

### 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,