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

Question

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

Answer 1

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

Answer 2

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.

Answer 3

$x→0$ 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 $(0,0)$.