Multivariable limit involving sin and log

Question

I really can't solve this simple limit:

$$ \lim_{(x,y) \to (0,0)}\frac{\sin(x^2+y^2)}{\ln(3x^2y^2+1)} $$

Answer 1

The function is not defined if $x=0$ or $y=0$, because of a division by $0$. But let's suppose we exclude those points, and ask for a limit as $x,y \to 0,0$ excluding the axes. Then the limit is $+\infty$

In polar coordinates $x = r \cos(\theta)$ and $y = r \sin(\theta)$, $$ f(x,y) = \dfrac{\sin(r^2)}{\ln\left(1 + \frac{3}{8} r^4 (1 - \cos(4\theta))\right)} $$ The least this can be for fixed $r$ is when $\cos(4\theta) = -1$ (e.g. $\theta = \pi/4$, when we have $$ f(r/\sqrt{2},r/\sqrt{2}) = \frac{\sin(r^2)}{\ln(1 + 3 r^4/4)}$$ As $r \to 0$, $\sin(r^2) \sim r^2$ while $\ln(1+3 r^4/4) \sim 3 r^4/4$. Thus for $r > 0$ sufficiently small, we have $\sin(r^2) > r^2/2$ and $\ln(1+3 r^4/4) < r^4/2$, so with $r = \sqrt{x^2+y^2}$, $$ f(x,y) \ge f(r/\sqrt{2},r/\sqrt{2}) \ge 1/r^2$$