what is the most efficient way for multi-variate limit problems If  we have something like:
$\lim_{(x,y)\implies (0,0)}\frac{x^2 y}{x^2+y}$
Here I tried to using polar coordinates such as $x=r\cos\theta$and $ y=r\sin\theta$ where $r$ converges to $0+$.
However this did not work here.
Also, I tried other substitutions that also failed,I am not sure how to figure out the right one for this problem.
So, my main question here is what are the main methods of checking if the limit exists for problems similar to the one above? Any advice is greatly appreciated.
 A: It doesnt limit to $0$. Let 
$$y=x^4-x^2$$ then 
$$\frac{x^2y}{x^2+y}=\frac{x^2(x^4-x^2)}{x^4}=x^2-1\to -1$$.
A: Let $f(x,y) = {\large{\frac{x^2 y}{x^2+y}}}$.

If we let $(x,y)$ approach $(0,0)$ along the line $y=0$, the limit of $f$ along that path is zero.

Note that for $|x|\ne1$,
\begin{align*}
&x^2y = x^2 + y\\[4pt]
\iff\;&y=\frac{x^2}{x^2-1}\\[4pt]
\end{align*}
hence, if we let $(x,y)$ approach $(0,0)$ along the curve $y={\large{\frac{x^2}{x^2-1}}}$, with $|x|\ne 1$, the numerator and denominator of $f$ are equal and nonzero, so the limit of $f$ along that path is $1$. 

Hence, $f$ does not have a limit, as $(x,y)$ approaches $(0,0)$.

For the general problem of this type, you have to experiment.

If the limit doesn't exist, the usual strategy is find either


*

*A path, in the domain of $f$, approaching $(0,0)$, such that, as $(x,y)$ approaches $(0,0)$ along that path, the limit of $f$ does not exist.$\\[4pt]$

*Two paths, in the domain of $f$, approaching $(0,0)$, such that, as $(x,y)$ approaches $(0,0)$ along each path, the limit of $f$ exists, but the two limits are not the same.
 

Finding such paths is an art, but experience helps. 

A typical strategy would be to let $(x,y)$ approach $(0,0)$ along a path (in the domain of $f$) of the form
$$x=at^p,\;\;\;y=bt^q$$
as $t$ approaches $0$ from above, where $a,b,p,q$ are fixed real numbers, with $a,b$ not both zero, and $p,q > 0$. 

Then try to find two $4$-tuples $(a,b,p,q)$, where the limits differ (or where at least one of them doesn't exist).

Linear paths makes sense as a first try, and if those don't resolve the problem, parabolic paths are a natural next try.

For example, 


*

*If $(a,b,p,q)=(1,b,1,1)$, with $b\ne 0$, the path is approaching $(0,0)$ along the line $y=bx$, with $x$ approaching zero from above.

*If $(a,b,p,q)=(1,b,1,2)$, with $b\ne 0$, the path is approaching $(0,0)$ along the parabolic path $y=bx^2$, with $x$ approaching zero from above.



Of course, if the limit does exist, then you won't be able to find paths along which $f$ has different limits. 

In that case, try evaluating $f$ on a family of simple closed which "shrink" to $(0,0)$.

For example, consider the values of $f$ on a circle of radius $r$, centered at $(0,0)$. Assuming $f$ is continuous on each circle, let $m_r,M_r$ be the minimum and maximum values of $f$, restricted to that circle. If you can show that $m_r$ and $M_r$ approach a common limit, $L$ say, as $r$ approaches zero from above, then that proves that $f$ approaches $L$ as $(x,y)$ approaches $(0,0)$. This is equivalent to evaluating the limit of $f$, as $r$ approaches zero from above, after changing $(x,y)$ to $(r\cos(\theta),r\sin(\theta))$, (i.e., changing to polar coordinates).

As another example, for $a > 0$. you can use the family of axis-aligned squares, centered at $(0,0)$ with side length $2a$, defined algebraically as
$$\{(x,y)\mid \max(|x|,|y|)=a\}$$
Then as before, assuming $f$ is continuous on each square, let $m_a,M_a$ be the minimum and maximum values of $f$, restricted to that square. If you can show that $m_a$ and $M_a$ approach a common limit, $L$ say, as $a$ approaches zero from above, then that proves that $f$ approaches $L$ as $(x,y)$ approaches $(0,0)$.

As an alternative to using a family of simple closed curves shrinking to $(0,0)$, you can always resort to an epsilon-delta type proof, although in essence, it's the same thing.

In any case, at the outset, you need to get a sense of whether or not the limit exists, and if so, to what value. 

Unless it's algebraically obvious that the limit exists, I would always try a few paths first.

If the paths tried yield the same limit, then try instead to prove the limit of $f$ is equal to the limit found for those paths.
A: One consideration is that the denominator in your example is $0$ along a curve through $(0,0),$ namely $y=-x^2.$ And along this curve, the numerator doesn't vanish. Anytime this happens, the limit won't exist. This is useful to know.
