How to prove that $\lim\limits_{x,y\to 0} \frac{1}{x^2+y^2}$ does exist and what's its result?

Question

How to prove that the following limit doesn't exist ?

$$\lim\limits_{x,y\to 0} \frac{1}{x^2+y^2}$$

Answer 1

We will show that $$\lim\limits_{x,y\to 0} \frac{1}{x^2+y^2} = \infty$$

Remember that $\lim_{x \to a} f(x) = \infty \iff \forall_{\epsilon > 0} \exists_{\delta > 0}: |x - a| < \delta \Rightarrow f(x) > \frac1\epsilon$

That is, the limit of $f(x)$ goes to infinity when $x$ goes to $a$ iff, after you set a very big target number $T$, you can make $f(x)$ bigger than that target $T$ by making $x$ close to $a$. In the formal definition we instead choose a very small $\epsilon$, making $\frac1\epsilon$ very big.

Let us set some $\epsilon$. We need to show that we can make

$$\frac{1}{x^2+y^2} > \frac1\epsilon$$

by making $x, y$ very close to 0.

By commodity, we can assume that $x = y$ and then we get

$$\frac{1}{x^2+x^2} = \frac{1}{2x^2} > \frac1\epsilon$$

We can easily manipulate the expression getting

$$\frac{1}{2x^2} > \frac1\epsilon \iff\\ \frac{1}{x^2} > \frac2\epsilon \iff\\ x^2 < \frac\epsilon2 \iff\\ x < \sqrt{\frac\epsilon2}$$

That is, if $x, y < \sqrt{\frac\epsilon2}$, then $\frac{1}{x^2+y^2}$ will be greater than $\frac1\epsilon$ and thus we can make it infinitely big by making $x, y$ infinitely small and we established that

$$\lim\limits_{x,y\to 0} \frac{1}{x^2+y^2} = \infty$$

Answer 2

First step is to determine if the function is continuous near the limit.

We know that $\frac{1}{x^2+y^2}$ is only discontinuous at $(0,0)$. Since we are evaluating the function no where near that the path to taking the limit doesn't matter.

I'll choose (as @RSerraro did) the path of $||x^2+y^2||\rightarrow \infty$.Thus: \begin{equation} \lim_{x,y\rightarrow\infty}\frac{1}{x^2+y^2}=\lim_{||x+y||^2\rightarrow\infty}\frac{1}{x^2+y^2} = \lim_{x\rightarrow\infty}\frac{1}{x^2}=0 \end{equation}

For the limit as $(x,y)\rightarrow 0$:

I will note that I can reparamterize this function as:$\frac{1}{r^2}$

I notice that the function is independent of angle of approach so the limit can be rewritten as purely a function of $r$:

\begin{equation} \lim_{r\rightarrow 0,\theta \rightarrow ??} \frac{1}{r^2} = \lim_{r\rightarrow 0} \frac{1}{r^2} = \infty \end{equation}