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

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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.

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.

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.

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.