Finding $\lim_{(x, y)\to (0, 0)} \frac{(x^5−y^7)}{(x^4+y^4)}.$

Question

$$f(x,y)=\frac{(x^5 - y^7)}{(x^4+y^4)}$$ I have find its limit as $(x,y)\to (0,0).$ The answer is $0.$

I tried transforming it to polar coordinates, I got rid of the denominator but the numerator is tricky and I don't know how to proceed from there.

I also thought about using the squeeze theorem, but then again I can't find what value it should be greater and less than be.

edit: Polar coordinates: let $x = r\cos(a), \;y = r\sin(a)$.

After simplifying the function I get

$r^5\cos^5(a) - r^2\sin^7(a)$

Answer 1

$$|x^5| \leq |x| (x^4 + y^4)$$ $$|y^7| \leq |y^3| (x^4 + y^4)$$ $$ \left| \frac{x^5 - y^7}{x^4 + y^4} \right| \leq |x| + |y^3| $$

Answer 2

With polar coordinates you obtain $$\frac{x^5-y^7}{x^4+y^4}=\frac{r^5(\cos^5\theta-r^2\sin^7\theta)-}{(x^2+y^2)^2-2x^2y^2}=\frac{r^5(\cos^5\theta-r^2\sin^7\theta)}{r^4-2r^4\sin^2\theta\cos^2\theta}=\frac{r(\cos^5\theta-r^2\sin^7\theta)}{1-\frac12\sin^22\theta}$$ Now, if $r\le 1$, the numerator is bounded from above by $2r$, and the denominator is bounded from below by $\frac12$, so that $$0\le\Biggl|\frac{x^5-y^7}{x^4+y^4}\Biggr|\le 4r.$$

Answer 3

I don't think polar coordinates is all that useful here. Just note

$$|f(x,y)| \le |x|\frac{x^4}{x^4+y^4} + |y|^3\frac{y^4}{x^4+y^4} \le |x|+|y|^3.$$

As $(x,y)\to (0,0),$ both $|x|$ and $|y|^3$ go to zero, hence so does their sum. By the squeeze theorem, the desired limit is $0.$