I can't solve this problem and I would be glad to get some help: Is the function $f(x,y) = x\ln(x^2 + 3y^2)$ continuous at $(0,0)$ $f(0,0)=0$?
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5$\begingroup$ I think you have to provide $f(0,0)$ in order to get an answer to this question. $\endgroup$ – Jan Jul 13 '19 at 20:49
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$\begingroup$ yes, thank you I added it to the post. $\endgroup$ – Sapir Shuker Jul 14 '19 at 13:15
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You didn't say what the value of the function is at $(0,0)$, but if $f(0,0) = 0$ then the function is continuous at $(0,0)$. Here are some hints on why this is the case.
Useful fact $1)$: $$|x\ln(x^2 + 3y^2)| \leq |(x^2 + 3y^2)^{1 \over 2}\ln(x^2 + 3y^2)|$$ Useful fact $2)$: $$\lim_{r \rightarrow 0^+}r^{1 \over 2} \ln r = 0$$ Useful fact $3)$: The squeeze rule.
Now try to put these all together, or use variations of the above statements.
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Hint:
Passing to polar coordinates gives $$|f(r,\phi)| = |r\cos(\phi)\cdot \ln(r^2(1+2\cos^2\phi))| \le r|\ln (r^2)| \xrightarrow{r\to 0} 0$$
since $r\mapsto\left|\ln r\right|$ is decreasing for small $r$ and $1+2\cos^2\phi \ge 1$.
Therefore $\lim_{(x,y) \to (0,0)} f(x,y) = 0$.
answered Jul 13 '19 at 21:51
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$\begingroup$ @TedShifrin It shows how to find the right family of paths one should consider. By passing to polar coordinates it is clear that the first term goes to $0$ when $r \to 0$, and that by picking $r =\frac{c}{\ln(\sin(2\phi))}$ the second term goes to different values. $\endgroup$ – mechanodroid Jul 13 '19 at 22:11
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1$\begingroup$ I disagree. It's just as natural to think about choosing $y=f(x)$ so that $x\ln f(x)$ has an arbitrary limit. I might agree that you would win if we had $\ln(x^2+y^2)$ ... :) $\endgroup$ – Ted Shifrin Jul 13 '19 at 22:15
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$\begingroup$ @TedShifrin Well, answers to questions like these which involve polar coordinates are in general not to be taken very seriously anyway, as there are situations like these: math.stackexchange.com/q/753381/144766 You are right, of course, this answer is mostly just for fun. $\endgroup$ – mechanodroid Jul 13 '19 at 22:16
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1$\begingroup$ Well, because @herb steinberg convinced me that you and I are both wrong, I looked more carefully at your solution. First, how did you get $1+2\cos^2\phi = \sin 2\phi$? Next, as $r\to 0$, $e^{c/r}$ is certainly not in the domain of $\arcsin$. $\endgroup$ – Ted Shifrin Jul 13 '19 at 22:57
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$\begingroup$ @TedShifrin It was completely wrong, thanks. This hopefully shows that the limit of $f$ at $(0,0)$ is $0$, modulo the situation in the link. $\endgroup$ – mechanodroid Jul 13 '19 at 23:08
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$x→0$ faster than the ln term →−∞, so the $f(x,y)$ is continuous at $(0,0)$ (proof follows). In fact $f(0,y)=0$ for any value of $y$. In case there is any uncertainty about $f(0,0)$ being undefined, just call it $=0$ at that point. .
Without loss of generality assume $x^2+3y^2\lt 1$
For $x\gt 0$, $xln(x^2)\lt xln(x^2+3y^2)\lt 0$.
For $x\lt 0$, $xln(x^2)\gt xln(x^2+3y^2)\gt 0$. In both cases $xlnx^2\to 0$ as $x\to0$, so that $f(x,y)\to 0$.
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Therefore $f(x,y)$ is continuous at $(0,0)$.
answered Jul 14 '19 at 17:06
