Is this proof of limit of $f(x,y)$ correct?

Question

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 $\lim\limits_{(x,y)\to(0,0)}\dfrac{\log\left(\dfrac{1}{\sqrt{x^2+y^2}}\right)}{\dfrac{1}{\sqrt{x^2+y^2}}}=0$.

Answer 1

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.

Answer 2

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, $$\lim\limits_{(x,y)\to(0,0)}\dfrac{\log\left(\dfrac{1}{\sqrt{x^2+y^2}}\right)}{\dfrac{1}{\sqrt{x^2+y^2}}}=0$$

Answer 3

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.