Is this proof of limit of $f(x,y)$ correct? The limit is this: 
$$\lim\limits_{(x,y)\to(0,0)}\dfrac{\log\left(\dfrac{1}{\sqrt{x^2+y^2}}\right)}{\dfrac{1}{\sqrt{x^2+y^2}}}.$$
I think I did the proof well. Using that
$\lim\limits_{z\to\infty}\dfrac{\log(z)}{z}=0$
$,\lim\limits_{t\to 0^+}\dfrac{1}{t}=\infty$
$,\lim\limits_{(x,y)\to(0,0)}\sqrt{x^2+y^2}=0,$
$ \sqrt{x^2+y^2}>0 \space$ and the characterization of limits with sequences, we obtain that $\lim\limits_{(x,y)\to(0,0)}\dfrac{\log\left(\dfrac{1}{\sqrt{x^2+y^2}}\right)}{\dfrac{1}{\sqrt{x^2+y^2}}}=0$. 
 A: Yes that's a fine proof, as an alternative since $\log x \le \sqrt x$ we have
$$\dfrac{\log\left(\dfrac{1}{\sqrt{x^2+y^2}}\right)}{\dfrac{1}{\sqrt{x^2+y^2}}} \le \dfrac{\sqrt{x^2+y^2}}{\sqrt[4]{x^2+y^2}}=\sqrt[4]{x^2+y^2} \to 0$$
and we can conclude by squeeze theorem.
A: Take, $x=r cos(\theta) , y = r sin(\theta) $
Now, we have, $$\lim_{r \to 0 } \frac{\log(\frac{1}{r})}{\frac{1}{r}} = 0 $$
So, the result,
$$\lim\limits_{(x,y)\to(0,0)}\dfrac{\log\left(\dfrac{1}{\sqrt{x^2+y^2}}\right)}{\dfrac{1}{\sqrt{x^2+y^2}}}=0$$
A: As we have $x^2+y^2$ in our equation, it is best to convert to polar.
Let $\sqrt{x^2+y^2}=r$. Note I didn't give the individual equations of $x$ or $y$, unlike A Learner's answer, because we don't have any other expression with $x$ or $y$.
Let's start by setting our limit. Substituting $x=y=0$ into $\sqrt{x^2+y^2} = r$, $r=0$. Therefore, we have
$$\lim_{r \to 0} \frac{\log(\frac{1}{r})}{\frac{1}{r}} = \lim_{r \to 0} r\log(\frac{1}{r})$$.
By Squeeze Theorem and $\log(x)\leq\sqrt{x}$, $$\lim_{r \to 0} r\log(\frac{1}{r})\leq \lim_{r \to 0} \frac{r}{\sqrt{r}}= \lim_{r \to 0} \sqrt{r} = 0.$$
But overall, your proof is solid.
