4
$\begingroup$

I am stuck on proving that the limit $L = \lim\limits_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^4}$ does or does not exist (using the definition $\forall\epsilon\in\mathbb R_0^+:\exists\delta\in\mathbb R_0^+:\forall(x, y)\in\mathbb R^2-(0,0):||(x,y)||<\delta\Rightarrow|\frac{x^2y^2}{x^2+y^4}-L|<\epsilon$. I've found that if the limit exists, then $L$ must be equal to $0$ (by approaching the limit on several paths like $y=x$). I haven't found a path were the limit is not equal to 0, so I assume the limit does exist, but I can't find a way to proof the limit using the $\epsilon-\delta$ definition.

$\endgroup$
0
$\begingroup$

Just notice that $0 \leq \frac{x^2y^2}{x^2 + y^4} \leq \frac{x^2y^2}{x^2} = y^2$ since $y^4 \geq 0$. Now you just need to choose a $\delta > 0$ small enough (using the euclidic norm might help you the most) so that $\vert y^2 \vert < \epsilon$ and you are done :)

$\endgroup$
3
$\begingroup$

hint: $0 \le \dfrac{x^2y^2}{x^2+y^4} \le \dfrac{x^2y^2}{2|x|y^2} = \dfrac{|x|}{2}$. Use this double inequality together with the definition you can find the "$\delta$" in terms of "$\epsilon$" and completes the proof in elegant manner.

$\endgroup$
2
  • $\begingroup$ Where did you get $x^2+y^4\ge2\lvert x\rvert y^2$? $\endgroup$ – Akiva Weinberger Dec 30 '16 at 13:53
  • 1
    $\begingroup$ Because $(|x|-y^2)^2 ≥ 0$, so we get $x^2-2|x|y^2+y^4 ≥ 0$, then we find $x^2+y^4 ≥ 2|x|y^2$. $\endgroup$ – mitchbus Dec 30 '16 at 14:04
0
$\begingroup$

(This is @DeepSea 's answer, but without the zeros in the denominator)

Hint:

$$0\leq{x^2y^2\over x^2+y^4}={|x|\over2}\>{2|x|y^2\over x^2+y^4}\leq{|x|\over2}\ .$$

$\endgroup$

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.