2
$\begingroup$

I'd like to prove the following limit to be zero using the epsilon-delta definition: $\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{|x|^3+|y|^3}$

I tried to estimate |$\frac{x^2y^2}{|x|^3+|y|^3}$| to be $\lt\epsilon$ and thus find the fitting $\delta$. But I couldn't figure out the steps I need to take. I think the norm $\Vert.\Vert_3$ can be used here somehow. I can get rid of the $x^2y^2$ term like so:

Let $\epsilon\gt0$ and assume $\Vert X \Vert \lt1$, so $|x|\lt1$ and $|y|\lt1$.

Then $|\frac{x^2y^2}{|x|^3+|y|^3}-0| \lt |\frac{1}{(|x|^3+|y|^3)}|=\frac{1}{\Vert X \Vert_3}$

I would need to get the norm to the numerator and use $\Vert X \Vert_3 \le \Vert X \Vert_2$ to find $\delta$ so that $|\frac{x^2y^2}{|x|^3+|y|^3}-0| \lt \epsilon$ when $\Vert X-0 \Vert_2 \lt \delta$.

$\endgroup$
1
$\begingroup$

When one approach doesn't give you what you need, it is time to abandon it and try something else.

Divide the neighborhood of $0$ into two parts: where $|x| \le |y|$ and where $|y| \le |x|$. In the first region, divide top and bottom by $|y|^3$. Then the denominator is trapped between 1 and 2, which allows you to find bounds on the entire fraction. In the other region, divide by $|x|^3$ to do the same. Use the combined bounds to prove your theorem.

$\endgroup$
0
$\begingroup$

Since all norms are equivalent on $R^2$, there exist positive constants $c_1,c_2$ such that:

$$||(x,y)||_3\leq ||(x,y)||_2 \leq c_1||(x,y)||_3$$

and also

$$||(x,y)||_1\leq c_2||(x,y)||_2$$

where, as usual, $$||(x,y)||_p = \left(|x|^p+|y|^p\right)^{1/p}$$

Let $\epsilon > 0$. Take $\delta={\epsilon\over (c_1c_2)^4}$. Assume $||(x,y)||_2<\delta$. Then

$$|x||y|\leq (|x|+|y|)^2 =||(x,y)||_1^2\leq c_2^2||(x,y)||_2^2\leq c_2^2c_1^2||(x,y)||_3^2 $$ Squaring both sides: $$x^2y^2\leq (c_2c_1)^4||(x,y)||^4_3$$ whence, upon dividing by $||(x,y)||_3^3$, $${x^2y^2\over |x|^3+|y|^3}\leq (c_1c_2)^4||(x,y)||_3\leq(c_1c_2)^4||(x,y)||_2<(c_1c_2)^4\delta=\epsilon$$

thus proving that the limit is zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.