Does this limit exist? $\lim_{(x,y)\to (0,0)} \frac{x^2y^4}{(x^2+y^4)^2}$

Question

In a recent lecture I attended the following limit was discussed: $$\lim_{(x,y)\to (0,0)} \frac{x^2y^4}{(x^2+y^4)^2}$$ Multiple solutions were used to try and find the limit and illustrate how one would attack similar problems.

In particular we rewrote the expression using polar coordinates, but could however not conclude anything from it. When reviewing my notes i noticed some errors in this.

How should it be done, and what conclusions can be made from the results?

Another solution used was to substitute $x=t^2$ and $y=t$ resulting in the following expression: $$\lim_{t\to 0} \frac{t^4t^4}{(t^4+t^4)^2} = \frac{1}{4}.$$

Is this correct? How can I interpret the results?

Answer 1

Put $$y^2=Y$$ and $$x=r\cos(t)\; \; , Y=r\sin(t)$$

the limit becomes $$\lim_{r\to 0}\frac{r^4\cos^2(t)\sin^2(t)}{r^4}$$

which depends on the angle $t$

so, it does not exist.

Answer 2

If you approach along the $x$-axis $y=0,$ with $x\ne 0,$ you see that the limiting value is $0.$ However, as you have noted, travelling along the parabola $x=y^2$ gives $1/4.$ Thus, the limit cannot exist.

The reason for this is that for the limit to exist, then it must exist and have the same value however we approach the point in question. However, here, approaching the origin along two different paths gives us different results, so the limit does not exist.

For polar coordinates, this is the same, the limit must exist and must have the same value for all $\phi.$ However, as someone has demonstrated above, while the limit exists for all $\phi,$ it varies with $\phi.$ This is not acceptable by definition of the limit. So the limit does not exist, we conclude again.