Limits in functions from $\mathbb R^2 \to \mathbb R$ Good evening,
I would like tips / methods for calculating limits of functions of $\mathbb R ^ 2 \in \mathbb R$.
The squeeze theorem is sometimes a problem for me to set up because I do not know how to put in order this new type of functions in the good way, it is necessary that $f (x, y) <g (x, y) <h (x, y)$ but I must make sure this is true for all $x$ and all $y$?
Concerning the formal definition, making $\lvert \lvert x - x_0 \rvert \rvert$ (where $x$ and $x_0$ are in $\mathbb R^2$) appear in the expression $\lvert f(x) -l \rvert$ is sometimes difficult. Obviously, my teacher has the bad habit of giving the simplest examples in the world by looking for example at the limit when $(x, y)$ tends to $(1,0)$ of a function where there is already $\sqrt {(x-1) ^ 2 + y ^ 2}$ which is not very beneficial for the students.
I have heard of switching to polar. The course on these functions being discovered today, I have not yet seen this method but if it is really essential and useful I would like to know what it consists of.
As an example I am interested in the limit when (x, y) goes to $(0,1)$ of $f (x, y) = \frac {y} {x ^ 2} e ^ {\frac {- y} {x ^ 2}}$ for $x \neq 0$ and $0$ when $x = 0$.
 A: If you want to treat it as a single limit (as opposed to $x\rightarrow x_0$ and $y\rightarrow y_0$ independently) then you need to account for all possible sequences that converge to $(x_0,y_0)$. The limit of $f$ at $(x_0,y_0)$ only exists if the images of all such sequences converge to the same value.
This is because function limits are defined in terms of universal qualification of points arbitrarily close to the limit point. This doesn't make much of a difference in $\mathbb{R}$ because you can only approach a point from 2 different directions, but in higher dimensions there are infinite directions of approach.

If you know that the limit exists and you'd just like to compute it, then you can use any parametric function $z(n)\rightarrow(x_0,y_0)$, as long as $f$ is defined for every $z(n)$. Note that it's up to you how to use $n$; you could do $n\rightarrow\infty$ or $n\rightarrow0$. It just needs to match how you use $n$ in $z(n)$. ($n\rightarrow\infty$ is more common, such as in real-analysis proofs that use convergent sequences.)
You can also use this to prove that the limit doesn't exist, by creating a sequence where $f$ doesn't converge, or two sequences where $f$ converges to different values.
In your example, fixing $x=0$ is only one possible way to approach $(0,1)$. You also need to try other sequences as a sanity check, such as $\left(\frac{1}{n}\mathrm{cos}(n),1+\frac{1}{n}\mathrm{sin}(n)\right)$.
Note that there's nothing special about choosing the sequence, as long as it converges to $(x_0,y_0)$. The goal is to find a sequence that makes the computation of (or the proof of non-existence of) the limit convenient.
You also mentioned polar coordinates. If you used something like $\left(\frac{1}{n}\mathrm{cos}(\theta),\frac{1}{n}\mathrm{sin}(\theta)\right)$, then you can leave $\theta$ unspecified to show that the angle of approach doesn't affect the result.

Here is a specific example of a limit that appears to exist if you only look at one sequence approaching the limit point.
Let $f(x,y)=\mathrm{tanh}\left(\frac{1}{x+y}\right)$ and $(x_0,y_0)=(0,0)$.

*

*With $z(n)=\left(0,\frac{1}{n}\right)$, $f(z(n))=\mathrm{tanh}\left(n\right)\rightarrow 1$.

*With $z(n)=\left(\frac{1}{n},-\frac{2}{n}\right)$, $f(z(n))=\mathrm{tanh}\left(-n\right)\rightarrow -1$.

Since the images of the two sequences don't converge to the same value, that means that the limit at $(0,0)$ doesn't exist.
