Checking the continuity of function of 2 variables

Question

I have a function defined as$$f(x,y)=\frac{x^3+2y^3}{x^2+y^2}\,\forall\, (x,y)\neq (0,0)$$ $$f(x,y) = 0\,for\,(x,y)=(0,0)$$
I had to show that the given function is continuous at (0,0)

then the general idea is to show that $\lim_{(x,y)\to(0,0)}f(x,y)=l$ exists if for given $\epsilon<0\, \exists$ a $\delta>0$ such that $|f(x,y)-l|<\epsilon$ whenever $|x|<\delta$ and $|y|<\delta$

I started out with assuming $x=r\,cos\theta$ and $y=r\,sin\theta$

now using the above assumptions, I can write $$|f(x,y)-0|=|r(\cos^3\theta+2\sin^3\theta)|$$ Now I can take $r$ out of modulus function as it is always +ve and use the inequality that $|a+b|\le|a|+|b|$, this implies that above equation is equal to $$|r(\cos^3\theta+2\sin^3\theta)|\le r(|\cos^3\theta|+2|\sin^3\theta)|$$ this is where the problem lies; I don't understand how to reduce this form into a reasonable expression s.t. I could equate it to $\epsilon$ and define the $\delta$ that accompanies it

Answer 1

Hint: $r(|\cos^3\theta|+2|\sin^3\theta)|) \le r(1+2)=3r=3 \sqrt{x^2+y^2} \le 3(|x|+|y|)$

Answer 2

No need to get into $\epsilon$ and $\delta$. As you've stated, $$f(r\cos(\theta), r\sin(\theta)) = r(\cos^3(\theta)+2\sin^3(\theta))$$ Now, note that $\lvert \cos^3(\theta)+2\sin^3(\theta)\rvert\leq 3$ for $\theta\in [0, 2\pi)$. Therefore, $$\lvert f(r\cos(\theta), r\sin(\theta))\rvert\leq 3r$$ which implies that $$\lim_{r\to 0} f(r\cos(\theta), r\sin(\theta)) = 0$$ for all $\theta$ by the squeeze theorem.

Answer 3

You can also do like this:

You want to find a $\delta>0$ such that for all $\epsilon>0$ $$0<\sqrt{x^2+y^2}<\delta\implies |f(x,y)-0|<\epsilon$$ Now let $\delta=\epsilon/3$. $$\left|\frac{x^3+2y^3}{x^2+y^2}\right|\leq\left|\frac{x^3}{x^2+y^2}\right|+\left|\frac{2y^3}{x^2+y^2}\right|=|x|\left|\frac{x^2}{x^2+y^2}\right|+2|y|\left|\frac{y^2}{x^2+y^2}\right|$$ $$\leq|x|+2|y|\leq \sqrt{x^2+y^2}+2\sqrt{x^2+y^2}=3\sqrt{x^2+y^2}<3\delta=\epsilon.$$