What is the value of $\delta $ if $\epsilon=0.01$?

Question

Let $f(x,y) = \begin{cases} \frac{2x^2y+3xy^2}{x^2+y^2}, & \text{if $(x,y)\neq(0,0)$} \\[2ex] 0, & \text{if $(x,y)=(0,0)$ } \end{cases}$

Then the condition on $\delta $ such that $\vert f(x,y)-f(0,0) \vert<0.01$ whenever $\sqrt {x^2+y^2}<\delta $ is-

1.$\delta <0.01$

2.$\delta <0.001$

3.$\delta <0.02$

4.no such $\delta $ exists

solution:since $f(0,0)=0$,then consider $\left\vert \frac{2x^2y+3xy^2}{x^2+y^2}-0\right\vert =\left\vert \frac{xy(2x+3y)}{x^2+y^2}\right\vert\le \frac{(2x+3y)}{2}$ as $xy\le \frac{x^2+y^2}{2}$

From here, how to proceed further...

Answer 1

Note that we have

$$\frac12|2x+3y|\le |x|+\frac32|y|\le \frac52\sqrt{x^2+y^2}<0.01$$

whenever $\sqrt{x^2+y^2}<\delta=0.004$

Answer 2

To get a larger possible value of $\delta$, use the polar coordinate transformation $x = r \cos t, y = r \sin t$ to rewrite the equality in the question.

$$\left\vert \frac{2x^2y+3xy^2}{x^2+y^2}-0\right\vert =\left\vert \frac{r^3 \cos t \sin t (2\cos t+3\sin t)}{r^2}\right\vert = r |\cos t| |\sin t| |2 \cos t + 3 \sin t| $$

In another answer, choosing $|x|,|y| \le \sqrt{x^2+y^2}$ is similar to bounding $|\sin t|,|\cos t|$ by $1$. IMHO, it's too brutal as all information about the variable $t$ is lost. $2 \cos t + 3\sin t = \sqrt{2^2+3^2} \sin(t + \alpha) = \sqrt{13} \sin(t+\alpha)$, where $\tan \alpha = \dfrac23$.

From this

$$\left\vert \frac{2x^2y+3xy^2}{x^2+y^2}\right\vert = \frac{\sqrt{13}}{2} r |\sin 2t \sin(t+\alpha)| \le \frac{\sqrt{13}}{2} \delta < \epsilon$$

for any $t \in [0,2\pi]$ and $r < \delta$.

The choice $\delta < \dfrac{2}{\sqrt{13}} \epsilon$ guarantees that $|f(x,y)-f(0,0)| < \epsilon$. When $\epsilon = 0.01$, it becomes $$\delta < \frac{2}{\sqrt{13}} \cdot \frac{1}{100} = \frac{1}{50\sqrt{13}} \approx 0.005547. \quad \text{(cor. to 4 d.p.)}$$

This improves the bound given in another answer.

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The graph of $g(t) = \cos t \sin t (2 \cos t + 3 \sin t)$

Observe that the amplitude is between $1.5$ and $2$. This suggest that choosing $\dfrac52 \delta < \epsilon$ is too restrictive. We can allow more points to enter this $\delta$-ball in the domain of $f$ by choosing a larger $\delta$. For a fixed $\epsilon$, this means choosing a smaller coefficient of $\delta$. The choice $\dfrac{\sqrt{13}}{2} \approx 1.8028$ is nearer to the amplitude than $2.5$.

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Another answer shows that (2) "$\delta < 0.001$" is true and (4) "no such $\delta$ exists" is false. To complete this question, it remains to discuss the possibility of (1) and (3). To illustrate the sharpness of the bound $\delta < \dfrac{1}{50 \sqrt{13}}$ given above, choose $(r,t) = \left(0.0057, \dfrac{13}{50} \pi\right)$. This gives $(x,y) \approx (0.0039, 0.0042)$, and $f(x,y) \approx 0.010114 > 0.01 = \epsilon$.

Hence, (1) and (3) do not hold.