Computing A Multivariable Limit: $\lim_{(x,y) \to (0,0)}\frac{2x^2y}{x^4 + y^2}.$ Please help me in computing the following limit.
$$\lim_{(x,y) \to (0,0)}\frac{2x^2y}{x^4 + y^2}.$$
This will be my first attempt in computing a limit involving 2 variables.
Is this a part of multivariable calculus as it contains more than one variable?
How can I interpret this geometrically?
Thank You
 A: The reason we teach this particular problem is to show that directional limits along straight lines are not enough. Along a straight line $y=mx,$ with $m \neq 0,$ we have
$$ f(x,mx) = \frac{2 x^2 y}{x^4 + y^2} = \frac{2 m x^3 }{x^4 + m^2 x^2} = \frac{2 m x }{x^2 + m^2}  $$ from which
$$ |f(x,mx) | = \left| \frac{2 m  }{x^2 + m^2} \right| \cdot |x| \leq \left| \frac{2 m  }{ m^2} \right| \cdot |x| = \left| \frac{2   }{ m} \right| \cdot |x|. $$ 
Also, if we take either a vertical line $x=0$ or a horizontal line $y=0$ we get 0.
So, approaching the origin along any straight line gives an evident limit of 0.  In one variable, that would be enough, but in at least two variables, that is not sufficient to show that there is a limit, just approach the origin along a parabola $y = m x^2$ instead, as in Davide's answer.
A: As you have pointed out: Putting $(x,y) = (t,t^2)$ will give you:
$$\lim_{t\to 0} \frac{2t^4}{2t^4} = \lim_{t\to 0}1 = 1$$
Using the path $(x,y) = (t,0)$ gives
$$\lim_{t\to 0} \frac{ 2t^2 \cdot 0 }{t^4} = \lim_{t\to 0} 0 = 0$$
Since you got a different value in each case, the original limit cannot exist.
A: Put $f(x,y):=\frac{2x^2y}{x^4 + y^2}$. Fix a real number $m$. Then for $x\neq 0$
$$f(x,mx^2)=\frac{2x^2mx^2}{x^4+m^2x^4}=\frac{2m}{1+m^2},$$
so if there was a limit, it would be $\frac{2m}{1+m^2}$. These one depends on $m$, which is absurd since the limit would be unique.
A: Another approach is to consider the change of variables, $x^2=r \cos \theta$ and $y=r \sin \theta$ then $\frac{2x^2y}{x^4+y^2}=\frac{2r^2 \cos \theta \sin \theta}{r^2}=2\cos \theta \sin \theta$ which depends on $\theta$ and hence the limit doesn't exist since it's not unique. 
A: A friend teaching an elementary course recalled having seen an example of a two variable limit where the approach along any line lead to the same value, but approach along a parabola failed and asked me what it was. Here's one way of reverse engineering this, thinking in terms of something like homogeneous coordinates.

Students at this point have seen the "new type of discontinuity" (that did not exist in single variable calculus) of the function
$$
f(x,y)=\frac{xy}{x^2+y^2}.
$$
Here approaching the origin radially from different directions leads to distinct values, typically found by converting to polar coordinates. With homogeneous coordinates we can make the crucial observation that $f$ really is a function of the variable $t=y/x$. Namely, if $x\neq0$ then
$$
f(x,y)=\frac{xy/x^2}{(x^2+y^2)/x^2}=\frac{y/x}{1+(y/x)^2}=\frac{t}{1+t^2}.
$$
From this the claim is obvious because the variable $t$ takes a constant value on any ray emanating from the origin, but varies wildly, because near the origin $t$ can take any value we choose.

How to modify this to meet the present needs? If we want the value to be a constant along a parabola of the form $y=ax^2$ a natural idea is to write everything in terms of, not the ratio $t$, but the ratio $u=y/x^2$. Any function $g(u)$ will then have different value $g(a)$ on parabolas $y=ax^2$ with different choices of $a$. But, we also have the secondary requirement of having the same limit along any line through the origin, i.e. one with constant $t=y/x$. Let's see. We have $u=t/x$, and we are approaching the origin radially. If $t\neq0$ then $x\to0$ and $u\to\pm\infty$ according to the signs of $t$ and $x$. Also, along the line defined by $t=0$ (= the $x$-axis) we have $u=0$.

The conclusion is that if we have a function $g(u)$ such that
  $$g(0)=\lim_{u\to\pm\infty}g(u),$$
  then we are in business.

All that remains is to observe that we can reuse the function $f(t)=t/(1+t^2)$ from above, and end up with
$$
g(x,y)=g(u)=\frac{u}{1+u^2}=\frac{y/x^2}{1+y^2/x^4}=\frac{x^2y}{x^4+y^2}.
$$
A: Let $f(x,y):=\frac{2x^2y}{x^4 + y^2}$.
Let $g(x,y):=\frac{1}{2}f(x,y)$.
Let $h(x,y):=\frac{xy}{x^2+y^2}$.

Lemma 1:
$\lim_{(x,y) \to (0,0)}h(x,y)$ doesn't exist.


Proof of Lemma 1:
Assume that $\lim_{(x,y) \to (0,0)}h(x,y)$ exists.
If $x>0$ and $y>0$, then $h(x,y)>0$.
If $x<0$ and $y>0$, then $h(x,y)<0$.
So, $\lim_{(x,y) \to (0,0)}h(x,y)$ must be $0$.
If $(x,y)=(r\cos\theta,r\sin\theta)$, then $h(x,y)=h(r\cos\theta,r\sin\theta)=\cos\theta\sin\theta=\frac{1}{2}\sin(2\theta).$
$\frac{1}{2}\sin(2\theta)$ is not equal to $0$ for some $\theta$.
So, $\lim_{(x,y) \to (0,0)}h(x,y)$ is not $0$.
This is a contradiction.
So, $\lim_{(x,y) \to (0,0)}h(x,y)$ doesn't exist.

If $\lim_{(x,y) \to (0,0)}f(x,y)$ exists, then $\lim_{(x,y) \to (0,0)}g(x,y)$ exists.
So, if $\lim_{(x,y) \to (0,0)}g(x,y)$ doesn't exist, then $\lim_{(x,y) \to (0,0)}f(x,y)$ doesn't exist.
We prove that $\lim_{(x,y) \to (0,0)}g(x,y)$ doesn't exist.
Assume that $\lim_{(x,y) \to (0,0)}g(x,y)$ exist.
If $x\neq 0$ and $y>0$, then $g(x,y)>0$.
If $x\neq 0$ and $y<0$, then $g(x,y)<0$.
So, $\lim_{(x,y) \to (0,0)}g(x,y)$ must be $0$.
Let $\varepsilon$ be any positive real number.
Then there exists a positive real number $\delta$ such that $$0<\sqrt{x^2+y^2}<\delta\implies |g(x,y)|<\varepsilon.$$
Let $\delta^{'}:=\min(\frac{\delta^2}{2},1)$.
Let $(x,y)$ be an element of $\mathbb{R}^2$ such that $0<\sqrt{x^2+y^2}<\delta^{'}$.

*

*We consider the case in which $x\geq 0$.
$x\leq\sqrt{x^2+y^2}<\delta^{'}.$
$|y|\leq\sqrt{x^2+y^2}<\delta^{'}.$
$0<\sqrt{\sqrt{x}^2+y^2}=\sqrt{x+y^2}<\sqrt{\delta^{'}+(\delta^{'})^2}\leq\sqrt{2\delta^{'}}\leq\sqrt{2\frac{\delta^2}{2}}=\delta.$
So, $|h(x,y)|=|g(\sqrt{x},y)|<\varepsilon.$

*We consider the case in which $x\leq 0$.
$-x\leq\sqrt{x^2+y^2}<\delta^{'}.$
$|y|\leq\sqrt{x^2+y^2}<\delta^{'}.$
$0<\sqrt{\sqrt{-x}^2+y^2}=\sqrt{-x+y^2}<\sqrt{\delta^{'}+(\delta^{'})^2}\leq\sqrt{2\delta^{'}}\leq\sqrt{2\frac{\delta^2}{2}}=\delta.$
So, $|h(x,y)|=|-h(-x,y)|=|-g(\sqrt{-x},y)|=|g(\sqrt{-x},y)|<\varepsilon.$
So, $\lim_{(x,y) \to (0,0)}h(x,y)=0.$
By Lemma 1, this is a contradiction.
So, $\lim_{(x,y) \to (0,0)}g(x,y)$ doesn't exist.
