Limits and Continuity of Functions of Two Variables

I have a problem with the following function:

f(x, y)=

{xy(x^2 - y^4)/(x^4 + y^2) if (x, y) =/= (0, 0)
{
{0 if (x, y)=(0, 0)

I have to prove that f is continuous in (0, 0).

Is it possible to show the existence without using polar coordinates?

Help me please :)

Thank you so much!

Here's a δ-ε proof of this.

To assure that (x, y) is sufficiently close to (0, 0), I'll assume that √(x² + y²) < 1.
In particular, we have |x|, |y| < 1, which I'll use repeatedly below.
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Note that |f(x, y) - f(0, 0)|
= |xy(x² - y⁴)/(x⁴ + y²) - 0|
= |xy(x² - y⁴)| / (x⁴ + y²)
≤ |xy| (|x²| + |y⁴|) / (x⁴ + y²), by triangle inequality
= |xy| [x²/(x⁴ + y²) + y⁴/(x⁴ + y²)]
< |xy| [x²/(x⁴ + x²y²) + y⁴/(0 + y²)], decreasing denominators
= |x| |y| [1/(x² + y²) + y²]
< |x| |y| [1/√(x² + y²) + y²], since √(x² + y²) < 1
= √x² √y² [1/√(x² + y²) + (√y²)⁴]
≤ √(x²+y²) √(x²+y²) [1/√(x² + y²) + √(x² + y²)⁴]
= √(x² + y²) + √(x² + y²)⁵
< √(x² + y²) + √(x² + y²), since √(x² + y²) < 1
= 2√(x² + y²).

Finally, given ε > 0, let δ = min{1, ε/2}.
Then, √(x² + y²) < δ ==> |f(x, y) - f(0, 0)| < 2√(x² + y²) < 2(ε/2) = ε, as required.

Hence, f is continuous at (0, 0).

I hope this helps!
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P.S.: The above can be converted into a squeeze law argument:
0 < |f(x, y)| < 2√(x² + y²) for all (x, y) sufficiently near (0, 0).

Since lim((x,y)→(0,0)) 2√(x² + y²) = 0, we conclude that lim((x,y)→(0,0)) |f(x,y)| = 0.

Hence, lim((x,y)→(0,0)) f(x,y) = 0 = f(0, 0), and we are done.

This was a tough one to do with polar coordinates!

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