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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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Limit does not exists, if you consider two paths of approach $y=0$ and $x=\sqrt{y}$. – Sangchul Lee Dec 24 '11 at 15:20
Try converting to polar coordinates. – J. M. Dec 24 '11 at 15:20
Would you be able to calculate the limit if you assume that $y=x^2$? Would you be able to do the same if you assume $y=x$? If these limits are different, it means that you can obtain different values if you approach $(0,0)$ along different curves and, consequently, the limit in two variables does not exist. – Martin Sleziak Dec 24 '11 at 15:20
In general, to show that the limit does not exist at a point, you may use "Two Path Test" which @sos440 used above. – HbCwiRoJDp Dec 24 '11 at 17:00
up vote 2 down vote accepted

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.

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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.

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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.

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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.

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