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 $(0,0)$ $f(0,0)=0$?
 A: 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.
A: 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 $\lim_{(x,y) \to (0,0)} f(x,y) = 0$.
A: $x→0$
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 $(0,0)$.
