Multivariable Calculus Limits

Question

$$\lim_{(x,y) \to (0,0)}= \frac{xy^3}{x^2-x^4 + y^4}$$

This is one of the limits that I have been tasked with calculating and I have tried searching similar limits on this website so I can learn how to solve this but I haven't found any... Also, I haven't been taught conversion to polar coordinates which might have made this easier.

My best guess so far has been to use the sandwich theorem with the $|xy^3|$ on either side but that computes it to be zero and I'm not very sure of this method.

$$\lim_{(x,y) \to (0,0)} \frac{x^3 + y^4}{x^2 + y^4}$$ And this is the second one:

Now I did find a similar question to this but the method and answer were vastly different. Here I am thinking of using the two-path method with $x=0$ and $x=y$ which shows that the limit diverges (DNE). Would that be correct?

Answer 1

First, let's restrict $x$ so that $|x|<1$. Then, $x^2-x^4>0$ and by the AM-GM inequality we have

$$y^4+(x^2-x^4)\ge 2y^2\sqrt{x^2-x^4}$$

Therefore, we assert that

$$\left|\frac{xy^3}{y^4+x^2-x^4}\right|\le \frac{|y|}{2\sqrt{1-x^2}}\to 0 \,\,\text{as}\,\,(x,y)\to (0,0)$$

Answer 2

Notice that $x^4=o(x^2)$ when $x\to 0$ so we can basically ignore the $-x^4$ term on the denominator.

Assuming we have $x^2+y^4$ then it is advantageous to set $u=\frac y{\sqrt{x}}$ .

After substitution we arrive to $f(x,y)=\dfrac{u^3\sqrt{x}}{1+u^4-x^2}$

Since $x\to 0$ we can assume $|x|<\frac 12$ and $|f(x,y)|<\underbrace{\left|\dfrac{u^3}{\frac 34+u^4}\right|}_\text{bounded}\times\underbrace{\sqrt{x}}_{\to 0}\to 0$

Rem: the term $|g(u)|$ is continuous (since denominator does not annulate) of limit zero at infinity, thus bounded.