Multivariable limit - perhaps a trickier problem I am stuck on. [duplicate]

Solution 1:

One may observe that, as $(x,y) \to (0,0)$, with $x>0$ and $y=\sqrt{x}$, one gets $$ \frac{x^4y^4}{(x^2 + y^4)^3}=\frac18, $$ on the other hand, as $(x,y) \to (0,0)$, with $y>0$ and $x=\sqrt{y}$, one gets $$ \frac{x^4y^4}{(x^2 + y^4)^3}=\frac{y^3}{\left(1+y^3\right)^3}\to 0, $$ thus the limit of the given function does not exist.

Solution 2:

If $$f(x,y) = \frac{x^4 y^4}{(x^2 + y^4)^3},$$ this suggests that when $x^2$ and $y^4$ tend to $0$ at the same rate, i.e., if we let $x^2 = y^4$, then we find $$f(y^2,y) = \frac{(y^2)^4 y^4}{((y^2)^2+y^4)^3} = \frac{y^{12}}{(2y^4)^3} = \frac{1}{8}.$$ This is the desired path that demonstrates the limit does not exist at $(0,0)$, since another choice such as along the line $(x,0)$ gives a limit of $0$.

Solution 3:

Use polar coordinates : $$\frac{x^4y^4}{(x^2+y^4)^3} = \frac{r^8\cos^4\theta\sin^4\theta}{r^6(\cos^2\theta+r^2\sin^4\theta)^3} = r^2\frac{\cos^4\theta\sin^4\theta}{(\cos^2\theta+r^2\sin^4\theta)^3}$$ Now the problem remaining is : is the fraction bounded or not ? You have, for $\theta\ne\frac\pi2\mod\pi$ : $$\frac{\cos^4\theta\sin^4\theta}{(\cos^2\theta+r^2\sin^4\theta)^3} = \frac{\cos^4\theta\sin^4\theta}{\cos^6\theta\left(1+r^2\frac{\sin^4\theta}{\cos^2\theta}\right)^4}$$ So if you go to $0$ following the curve of equation $r=\cos\theta$, for example (this is a circle tangent to the $y$ axis at the origin), you have : $$\frac{x^4y^4}{(x^2+y^4)^3} = \cos^2\theta\frac{\cos^4\theta\sin^4\theta}{\cos^6\theta(1+\sin^4\theta)}\xrightarrow[\theta\to\frac\pi2]{}\frac12\ne0$$ So your function is not continuous at the origin, as you suspected.