# Limit of $\lim_{(x,y)\rightarrow (0,0)}\ xy\log(\lvert x\rvert+\lvert y\rvert)$

I need help to understand how we compute this kind of limit:

$\lim_{(x,y)\rightarrow (0,0)}\ xy\log(\lvert x\rvert+\lvert y\rvert)$

I think we can use the squeeze theorem but I don't know how to bound the function, so I can use the theorem. If I suppose $0 \lt \sqrt{x^2+y^2} \lt 1$ then, but I'm struggle..

$0 \le \lvert f(x,y)\rvert = \lvert xy\log(\lvert x\rvert+\lvert y\rvert)\rvert$

Thanks in advance for the help.

• To format $xy\log(\text{foo})$, use xy \log(\text{foo}) Lower case $l$. May 29, 2018 at 22:03

Use the AM-GM inequality: $$0\leq 2\sqrt{\lvert xy\rvert} \leq |x|+|y| \tag{1}$$ from which $$0\leq 2\sqrt{\lvert xy\rvert}|\log(|x|+|y|)| \leq (|x|+|y|)|\log(|x|+|y|)|\tag{2}$$ Setting $t(x,y)\stackrel{\rm def}{=} |x|+|y|$, we have $\lim_{(x,y)\to(0,0)} t(x,y)=0$ (why?) and therefore since $\lim_{t\to 0}t\log t = 0$ we get by the Squeeze theorem

$$\lim_{(x,y)\to(0,0)}\sqrt{\lvert xy\rvert}|\log(|x|+|y|)| = 0\tag{3}$$ and also $$\lim_{(x,y)\to(0,0)}\sqrt{\lvert xy\rvert} = 0 \tag{4}$$ also by (1) and the Squeeze theorem. Can you conclude?

• Nice use of AM-GM!
– user
May 29, 2018 at 22:18
• Thank you! ${}{}$ May 29, 2018 at 22:22

HINT

We have that

$$xy\log(\lvert x\rvert+\lvert y\rvert)=\frac{xy}{|x|+|y|}\cdot (|x|+|y|)\log(\lvert x\rvert+\lvert y\rvert)$$

then recall that as $u\to 0$

• $u\log u \to 0$

and by polar coordinates

• $\frac{xy}{|x|+|y|}=r\cdot\frac{\cos \theta \sin \theta}{|\cos \theta|+|\sin \theta|}$

with

• $|\cos \theta|+|\sin \theta|\ge m>0$

Or as an alternative directly by polar coordinates

$$xy\log(\lvert x\rvert+\lvert y\rvert)=r\cos \theta \sin \theta\cdot \left(r\log r+r\log(|\cos \theta|+|\sin \theta|)\right)$$

• Gimusi i have a question. In case i try to solve a limit with polar coordinates, and the result is a constant. Can i assure the limit is that constant? May 29, 2018 at 22:10
• @TheNicouU Can you give an example for this case?
– user
May 29, 2018 at 22:14
• want me to ask a question with an example? It might be interesting. Plus I do not want to interfere with this thread. May 29, 2018 at 22:16
• @TheNicouU Yes it is of course the best way! In such way you can obtain more advices.
– user
May 29, 2018 at 22:17