# Limit of a multivariable function: $x^2\log(x^2+y^2)$, $(x,y)\to(0,0)$

Can someone help me to solve this limit?

$$\lim\limits_{(x,y)\to (0,0)} x^2\log(x^2+y^2)$$

For any line $y=mx$ the result is $0$, so, the candidate is $0$. I tried to use the squeeze theorem, polar coordinates, etc. but I can't solve it.

Thanks.

$0\le|x^2\log(x^2+y^2)|\le|(x^2+y^2)\log(x^2+y^2)|$. But $t\log(t)\to0$ when $t\to 0$

• That's a simple way to prove it. Thanks! – Nicolás Giossa Jan 11 '16 at 20:56
• @NicolásGiossa You still need to prove that $t\log t \to 0$ as $t \to 0^+$. Notice that $\log t \to -\infty$ as $t \to 0^+$. – Fly by Night Jan 12 '16 at 18:34
• @FlybyNight Sure, but it's easy using L'Hôpital's rule. – Nicolás Giossa Jan 12 '16 at 18:43
• @NicolásGiossa How would you apply L'H's rule to a product? – Fly by Night Jan 12 '16 at 19:11
• @NicolásGiossa $t\log t=\dfrac{\frac{\log t}{1}}{\frac{1}{t}}$ – sinbadh Jan 12 '16 at 19:13

Here is a way to do it with polar coordinates. Set $x=r\cos \theta$ and $y=r\sin\theta$. Then, you have $$\lim_{r\to 0} r^2\cos^2\theta \log(r^2)$$ The term $\cos^2 \theta$ is bounded, so if $r^2\log(r^2)$ goes to $0$, then the entire term above goes to $0$. We will apply l'Hopital's rule: $$\lim_{r\to 0}r^2\log(r^2)=\lim_{r\to 0}\frac{\log(r^2)}{{\frac{1}{r^2}}} \implies \lim_{r\to 0}-\frac{\frac{2r}{r^2}}{\frac{2}{r^3}}=\lim_{r\to 0}-\frac{2}{\frac{2}{r^2}}=\lim_{r \to 0}-r^2=0$$

HINT

I would use polar coordinates, i.e. $x=r\cos\theta$ and $y=r\sin\theta$. Looking at when $(x,y) \to (0,0)$ is the same as looking at when $r \to 0^+$.

If $x=r\cos\theta$ and $y=r\sin\theta$, and knowing that $\sin^2\theta + \cos^2\theta \equiv 1$, gives $$x^2\log(x^2+y^2) = r^2 \cdot \cos^2\theta \cdot \log (r^2)$$

We are then interested in the limit:

$$\cos^2\theta \left(\lim_{r \to 0^+} \ r^2 \cdot \log(r^2)\right)$$

Notice that if $r^2\cdot \log(r^2)$ has a non-zero limit then your original limit is undefined because it will depend on $\theta$. This is not allowed. It shouldn't matter what direction we approach $(0,0)$ from. We should get the same answer. In short: there are two possible answers:

• The limit is zero.
• The limit is undefined.

Since $r>0$ we can use laws of logs to give $$\lim_{r \to 0^+} \ r^2 \cdot \log(r^2)=2\lim_{r \to 0^+} \ r^2 \cdot \log(r)$$

Can you show that $r^2 \cdot \log(r) \to 0$ as $r\to 0^+$ ?

Observe

$$\lim\limits_{(x,y) \to 0}e^{x^2\log(x^2+y^2)}=\lim\limits_{(x,y) \to 0}(x^2+y^2)^{x^2}$$

Now for any $y_{0} < \infty$ including $y_{0} = 0$

$$\lim\limits_{x \to 0}(x^2+y_{0}^2)^{x^2}=1$$

(We may need to know the fact $\lim\limits_{t \to 0}t^t=1$.)

This means that $y \to 0$ part does not affect the limit and that it can be removed since $y$ can behave any way it wants.

$$\lim\limits_{(x,y) \to 0}(x^2+y^2)^{x^2}=\lim\limits_{x \to 0}(x^2 +|c|)^{x^2}=1=e^0$$

From $1=e^0$ it must be:

$$\lim\limits_{(x,y) \to 0}x^2\log(x^2+y^2)=0$$