Evaluate the continuity of this function

Question

Ok so after countless hours looking at video about epsilon-delta, I still can't understand it. I am asked to evaluate the continuity of this function using epsilon-delta.

$$f(x,y) = \begin{Bmatrix} \dfrac{5x^3+7y^4+6y^3-x^2y}{x^2+y^2} &if (x,y)\neq 0) \\0 &if (x,y)=0) \end{Bmatrix}$$

Does f is continuous at (0,0)? Explain

Thank you

Answer 1

Let $\varepsilon > 0$; then for all $(x,y) \neq (0,0)$ we have $$ \bigg| \frac{5x^{3} + 7y^{4} + 6y^{3} - x^{2}y}{x^{2}+y^{2}} \bigg| \leq \frac{5|x^{3}| + 7y^{4} + 6|y^{3}| + |x^{2}y|}{x^{2}+y^{2}}. $$ If $|x| \leq |y| \leq 1$, then $$ \frac{5|x^{3}| + 7y^{4} + 6|y^{3}| + |x^{2}y|}{x^{2}+y^{2}} \leq \frac{7y^{4} + 12|y^{3}|}{2x^{2}} \leq \frac{19|y^{3}|}{2x^{2}} \leq \frac{19}{2}|y|, $$ which is $< \varepsilon$ if in addition we have $|y| < 2\varepsilon/19$; therefore, if $|x| \leq |y| < \min \{ 1, 2\varepsilon/19 \}$, then the absolute value of the value of the map at $(x,y)$ can be made $< \varepsilon$.

Answer 2

Use polar coordinates: \begin{align*} \biggl\lvert\frac{5x^3+7y^4+6y^3-x^2y}{x^2+y^2}\biggr\rvert&=\frac{\bigl\lvert5r^3\cos^3\theta+7r^4\sin^4\theta+6r^3\sin^3\theta-r^3\cos^2\theta\sin\theta\bigr\rvert}{r^2}\\ &=r\bigl\lvert5\cos^3\theta+7r\sin^4\theta+6\sin^3\theta-\cos^2\theta\sin\theta\bigr\rvert\\ &\le r(5+7+6+1)=19r, \end{align*} if we suppose $r\le 1$. From there it is easy to use the $\varepsilon$-$\delta$ argument.