limit using polar coordinates of $f(x,y)$

Question

I have a doubt about finding the limit of two variable functions $f(x,y)$ at some general point say $(a,b)$ without loss of generality $a >0, b>0.$

The relation between cartesian coordinates and polar coordinates is as

$$x=r\cos(\theta), y=r\sin(\theta)$$ where $r \geq 0$ and $\theta \in (0, 2\pi].$

$$r=\sqrt{x^2+y^2}, \theta = \tan^{-1}\frac{y}{x}$$

Doubt: Do these two expressions are equivalent to finding the limit of $f(x,y)$?

$$ \lim_{(x,y)\to(a,b)}f(x,y)=\lim_{(r,\theta) \to (\sqrt {a^2+b^2},\theta)}f(r,\theta) -----(*)$$

where on the right-hand side in $(*)$, we fixed $r$ and then checked expression depends on $\theta $ or not we determine whether limit exists or not.

For instance limit at origin that is $(x,y) \to (0,0)$ then $(r,\theta) \to (0,\theta)$.

I'm a little confused about this concept. Please correct me if I'm wrong somewhere.

Thanks in advance.

Edit: For example

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

Using polar coordinates it is easily seen that $$\lim_{(r,\theta) \to (0,\theta)} r^2[\cos^2 \theta+r\sin^3 \theta/r\cos \theta+r\sin \theta+1]=0$$

The idea of polar coordinates, is used to simplify all the cases (generally infinite number) that come with two variable limits. If $y \to 0$, then $y = {x, x^2, x^3, \sin{x}, e^x - 1 \dots }$. You cannot evaluate all cases by hand in short amount of time. Since, $r$ has a direct relation to $x, y$, being $(\sqrt{x^2 + y^2})$, we can just let $r \to 0$ as the square root does not affect the limit approaching $0$. This reduces to one variable limit making it easier to solve, that being $\theta$ does not affect the limit (This is explained in the very first answer of the your first link) — Dstarred, Sep 08 '23 at 15:10
... Otherwise it is a two variable limit again and the polar method is ineffective. — Dstarred, Sep 08 '23 at 15:12

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