Why if any path going toward a point yields the same limit, then limit at that point exist? We have the definition of limit of multivariable function. Basically, for any given $t$, we can find a $q$, so that for any point, whose distance to the point where we want to calculate the limit is less than $q$, $t>|f-L|$. Then we say $L$ is its limit.
There is another way to state that the limit exists. That is, for any curve going toward the point, they have to yield the same limit.
So my question is: why is "any curve going toward the point has the same limit" equivalent to "limit exists"?
 A: This is a good question. I'll prove something slightly different: that if the limit $\lim_{x\rightarrow a}f(x)\not=L$ (maybe it exists, maybe not - just, it's not $L$), then there's some $\epsilon>0$ and some path going to $a$ such that the limit of $f$ along that path also isn't $L$ (whether or not it exists) - that is, there are points on the path arbitrarily close to $a$ where $f$ is at least $\epsilon$ away from $L$. It's not hard to show that this implies the result you're asking about.
So how do we do this? Suppose $\lim_{x\rightarrow a}f(x)\not=L$. Then for some $\epsilon>0$, there are points arbitrarily close to $a$ where $f$ gets at least $\epsilon$ away from $L$. We'll use these! For each $i\in\mathbb{N}$, let $b_i$ be some point within $2^{-i}$ of $a$, with $\vert f(b_i)-L\vert\ge\epsilon$. Now just "connect the dots" - consider the path going from $b_1$ to $b_2$ to $b_3$ to ... via straight lines. This path goes towards $a$, and the limit of $f$ along this path isn't $L$ (maybe it exists, maybe not - just, it's not $L$).
