# is the function $f(x,y) = x \ln(x^2 + 3y^2)$ continuous at $(0,0)$?

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 $$(0,0)$$ $$f(0,0)=0$$?

• I think you have to provide $f(0,0)$ in order to get an answer to this question. – Jan Jul 13 '19 at 20:49
• yes, thank you I added it to the post. – Sapir Shuker Jul 14 '19 at 13:15

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.

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 $$\lim_{(x,y) \to (0,0)} f(x,y) = 0$$.

• @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. – mechanodroid Jul 13 '19 at 22:11
• 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 $\ln(x^2+y^2)$ ... :) – Ted Shifrin Jul 13 '19 at 22:15
• @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. – mechanodroid Jul 13 '19 at 22:16
• 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 $\arcsin$. – Ted Shifrin Jul 13 '19 at 22:57
• @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. – mechanodroid Jul 13 '19 at 23:08

$$x→0$$ 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 $$f(x,y)\to 0$$.

.

Therefore $$f(x,y)$$ is continuous at $$(0,0)$$.