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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.

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.

jeea
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  • Your proof assumes both $x$ and $y$ got to $0$. It is easy to show that the limit is $0$ even if only one variable goes to $0$, but your proof does not apply there, since $r$ does not go to $0$. In this case you need to use $sin2\theta \to 0$, since $\theta \to 0$ or $\theta \to \frac{\pi}{2}$. – herb steinberg Aug 30 '18 at 03:54
  • @herb Sorry i dont understand why we use $\sin 2\theta \to 0$ – jeea Aug 30 '18 at 03:58
  • Of course $r\to 0$ here, while $\sin 2\theta$ does not even need to converge. I don't understand the comment made by @herbsteinberg. – Kusma Aug 30 '18 at 05:52
  • $r \to 0$ requires both $x$ and $y \to 0$. My comment was about the case where only one variable $\to 0$, but the other did not. In that situation $\theta \to 0$ or $\to \frac{\pi}{2}$, so $sin2\theta \to 0$. – herb steinberg Aug 30 '18 at 15:30
  • @jeea Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Sep 17 '18 at 20:27

3 Answers3

6

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!

Mark Viola
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2

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

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

user
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  • How to show $t\log t \to 0$ as $t\to 0$ ?? – jeea Sep 19 '18 at 07:40
  • It is a standard limit, to show that we can use l’Hopital (I don’t like) or (better) let $t=e^{-x}\to 0$ as $x\to \infty$ then $t\log t=-\frac x{e^x}\to 0$. – user Sep 19 '18 at 07:46
  • Again how to show $\frac{x}{e^x} \to 0$ ? I needed epsilon delta! – jeea Sep 19 '18 at 08:08
  • @jeea That also a standard limit, we can use again l'Hopita (I don't like) or (better) use that eventually $e^x>x^2 \implies \frac x{e^x}<\frac1x \to 0$. – user Sep 19 '18 at 08:12
  • @jeea Before solving multivariable limits you should have a strong base with those foundamental limits for functions of one variable and sequences, if you have doubts on that revise them. – user Sep 19 '18 at 08:13
  • a very good advice, I was out of time due to college examinations! – jeea Sep 19 '18 at 08:15
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    Here on MSE you can find easily the way to revise them! Bye – user Sep 19 '18 at 08:17
  • Of course, one could use $|\log(x^2+y^2)|\le \frac1{\sqrt{x^2+y^2}}$ and forgo appeal to the limit $\lim_{t\to0}t\log(t)=0$. – Mark Viola Jul 20 '20 at 18:15
1

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 $0^+$

Tsemo Aristide
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