<s>[INST] How to prove the following by epsilon delta definition? I have to prove that $$\underset{(x,y) \to (0,0)}{\lim} (x^2+y^2) \log(\sqrt{x^2+y^2}) = 0$$ by $\epsilon- \delta$ definition. I can easily prove it by L-Hospital rule, but unable to do with definition.
 [/INST] We need to prove that $\forall \epsilon>0,\exists \delta >0$ such that $\forall (x,y):\sqrt{x^2+y^2}<\delta \Rightarrow |(x^2+y^2) \log{\sqrt{x^2+y^2}}|< \epsilon$
Let $\epsilon>0$.
For $\delta \leq 1$ 
$$|(x^2+y^2) \log{\sqrt{x^2+y^2}}| \leq \frac{(x^2+y^2)^2}{2}+ \frac{(\log{\sqrt{x^2+y^2}})^2}{2} $$
Now $$\frac{(\log{\sqrt{x^2+y^2}})^2}{2}< \epsilon \Rightarrow \log{\sqrt{x^2+y^2}}<\sqrt{2 \epsilon} \Rightarrow \sqrt{x^2+y^2} <e^{\sqrt{2\epsilon}}$$
Take $\delta \leq\min\{1,2\sqrt[4]{\epsilon},e^{\sqrt{2 \epsilon}}\}$
</s>