Double limits with Polar coordinates

Question

We have to evaluate:

$$\lim_{(x,y) \to (0,0)} \frac{x^{4}-x^{2}y^{2}+y^{4}}{x^{2}+y^{2}+x^{4}y^{4}} \tag{1}$$

If we use polar coordinates and substitute $x=r\cos(\theta) $ and $y=r\sin(\theta)$ and take $r$ common from the expression in the numerator and that in the denominator and further do a bit of canceling we get that (1) is equivalent to,

$$\lim_{r \to 0}\frac{r^2(\cos^4\theta-\cos^2\theta \sin^2\theta + \sin^4\theta)}{\cos^2\theta+\sin^2\theta+r^6\cos^4\theta \sin^4\theta}$$

My teacher is saying that since the above expression has $r$ in front, therefore, the limit will go to zero. I don't think it is rigorous since we have $r$ in the denominator. Any suggestions. Can't seem to factorise these trigonometric expressions. Also tried $y=mx$ substitution to get different limits. They will all go to 0.

Answer 1

Another approach (I really dont like polar method)

$$\left|\frac{x^{4}-x^{2}y^{2}+y^{4}}{x^{2}+y^{2}+x^{4}y^{4}}\right|\le \frac{x^{4}+x^{2}y^{2}+y^{4}}{x^{2}+y^{2}+x^{4}y^{4}}\le $$ $$\le x^2 \underbrace{\frac{x^2+y^2}{{x^{2}+y^{2}+x^{4}y^{4}}}}_{\le 1}+y^2 \underbrace{\frac{y^2}{x^{2}+y^{2}+x^{4}y^{4}}}_{\le 1}\le x^2+y^2$$

So the Squezee Theorems guarantees that your limit is 0.

Answer 2

You're just forgetting that

• $r^2(\cos^4\theta-\cos^2\theta \sin^2\theta + \sin^4\theta)\le Ar^2$ for some $A$ since $0\le |\cos \theta|,|\sin\theta| \le 1$,
• $1+r^6\cos^4\theta\sin^4\theta\ge 1$. Therefore $$\frac{r^2(\cos^4\theta-\cos^2\theta \sin^2\theta + \sin^4\theta)}{1+r^6\cos^4\theta \sin^4\theta}\le\frac{Ar^2}1=Ar^2.$$

We simply use here that to get an upper bound of a fraction with positive numerator and denominator, we only need an upper bound for the numerator and a nonzero lower bound for the denominator.