Showing $\lim_{(x,y) \to (0,0)} xy \log(x^2+y^2) = 0$

Question

First I let $x=r\cos \theta, y = r\sin \theta$ and so limit

$$\lim_{r\to 0} r^2\sin2\theta \log(r)$$

Now, in region $0<x<1$, $\log(x) < 1/x$ $$|r^2\sin 2\theta \log (r) - 0| < |r\sin 2\theta| \le |r| < \delta < \epsilon$$

So limit exist if $\delta < \epsilon$ and limit is 0.

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Other way, I used L hospital, I don't know if we can apply, but I wrote $r^2 \log r$ as $\log(r) / (r^{-2})$ which again gave 0.

Answer 1

We have

$$xy \log(x^2+y^2) =(x^2+y^2)\log(x^2+y^2)\cdot \frac{xy}{x^2+y^2}\to 0$$

indeed since $t=x^2+y^2\to 0$

$$(x^2+y^2)\log(x^2+y^2)=t\log t\to 0$$

and since $x^2+y^2\ge 2xy$

$$0\le \left|\frac{xy}{x^2+y^2}\right| \le \frac12$$

Answer 2

More easily, $lim_{r\rightarrow 0}rlog(r)=0$, we deduce that $lim_{r\rightarrow 0}r^2log(r)=0$ since $|sin(\theta)|\leq 1$, the result follows.

Limit of $x \log x$ as $x$ tends to $0^+$

Answer 3

I thought it might be instructive to present an approach that is direct and follows from a pair of elementary inequalities only. To that end we proceed.

Using $\color{blue}{|\log(x)|<\frac1{\sqrt{x}}}$ for $x\le 1$ along with $\color{red}{|xy|\le \frac12(x^2+y^2)}$, we have for any given $\varepsilon>0$

$$\begin{align} \left|\color{red}{xy}\color{blue}{\log(x^2+y^2)}\right|&\le \color{red}{\frac12 (x^2+y^2)}\color{blue}{\frac1{\sqrt{x^2+y^2}}}\\\\ &=\frac12 \sqrt{x^2+y^2}\\\\ &<\varepsilon \end{align}$$

whenever $\sqrt{x^2+y^2}<\delta=2\varepsilon$.

And we are done!