Intuition behind polar coordinate to find limit in multivariate calculus

Question

Find limit using the polar coordinate for the function at $(0,0)$ $$ f(x,y) = \frac{x+y}{\sqrt{x^2+y^2}} $$

I started using $x = r\cos(\theta),\, y = r\sin(\theta)$

Then $(x,y) \to (0,0) \implies r \to 0$

Then we get as the following $$ f\bigl(r\cos(\theta),\, r\sin(\theta)\bigr) = \cos(\theta) + \sin(\theta) $$

Now I have no idea how to proceed from here. Some of the ideas from youtube videos I had was: $\theta$ is a free variable and the limit is more like spiraling into $(0,0)$.

Can someone explain to me why is this happening? What is the idea behind $r \to 0$?

Answer 1

If you want use polar coordinates, then with $(x,y)\mapsto (r\cos\theta,r\sin\theta)$ for $r\in {\bf R}_{+}^{*}$ and $\theta\in [0,2\pi[$, then $$f(r\cos\theta,r\sin\theta)=\frac{r(\cos\theta+\sin\theta)}{\sqrt{r^{2}(\cos^2\theta+\sin^2\theta)}}=\cos\theta+\sin\theta.$$

• If $\theta=\pi$, then $f_1(\theta):=f(r\cos\theta,r\sin\theta)=\cos\theta+\sin\theta=-1$.

• If $\theta=\pi/2$, then $f_2(\theta):=f(r\cos\theta,r\sin\theta)=\cos\theta+\sin\theta=1$.

If $f$ has limit $\ell\in {\bf R}$ at $(0,0)$, then the composite functions $f_1$ and $f_2$ they have to have the same limit $\ell$ at $0$, but it is imposible as it was show above. Therefore, the $\lim_{(x,y)\to (0,0)}\frac{x+y}{\sqrt{x^2+y^2}}$ does not exists.

Answer 2

The limit $\lim_\limits{(x,y)\to(0,0)} f(x,y)$

exists if $f$ evaluated every point in a small neighborhood of the origin has is in a small neighborhood. We can get more formal with $\epsilon-\delta$ balls, but this idea is sufficient.

If we convert to polar coordinate and then choose a small value of $r,$ then all values of $(r,\theta)$ will be close to each other. If the limit exists, all values of $f(r,\theta)$ will be close to each other regardless of our choice for $\theta.$

In this case $f(r,\theta) = \cos\theta + \sin \theta$

$f(r,\theta)$ could be as big a $\sqrt {2}$ and as small as $-\sqrt{2}$ no matter how small we choose $r$ to be. This is not a small neighborhood.