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I think the limit $$\lim\limits_{(x,y)\to(0,0)} \frac{xy^3}{x^2+y^4} $$ is equal to 0, but I have tried using polar coordinates and it only ends up multiplying the denominator by r (which goes to 0).

Other simple algebraic manipulations do not work. I think this could would using the epsilon-delta definition of the limit but I get $$\frac{|x||y|y^2}{x^2+y^4} $$ and I don't know how to go from that to the desired $\sqrt{x^2+y^2}$ .

Tora
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3 Answers3

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Hint:

$$x^2+y^4 \geq 2|x| y^2$$

By the AM-GM inequality . Or because $(|x|-y^2)^2 \geq 0$.

Now consider,

$$|\frac{xy^3}{x^2+y^4}|=\frac{|x||y|y^2}{x^2+y^4} \leq ...$$

Ahmed S. Attaalla
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I like doing these with Lagrange multipliers. For this one, with fixed $x^2 + y^4,$ we get the maximum value of $x y^3$ when $6 x^2 = 4 y^4,$ or $x = C y^2,$ with $|C| = \sqrt {\frac{2}{3}}$ Put that in the fraction, we get $$ \frac{C y^2 y^3}{C^2 y^4 + y^4} = \frac{Cy}{1 + C^2} $$ with limit zero.

Will Jagy
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We have that:

$$\left| \frac{xy^3}{x^2 + y^4} \right| = 2|x| y^2 \left| \frac{y}{2(x^2 + y^4)} \right| \leq \frac{x^2 + y^4}{2(x^2 + y^4)} |y| =\frac{1}{2} |y| \leq \sqrt{x^2 + y^2}$$

So, if we pick $\delta = \epsilon$, we have that $\sqrt{x^2 + y^2} < \epsilon$ and so $\left| \frac{xy^3}{x^2 + y^4} \right|$ gets arbitrarily close to $0$ as $(x, y)$ approaches $(0,0)$. Therefore the limit is equal to $0$.

Matheus Andrade
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  • Hmm how do you justify $| \dfrac {y^3}{x^2+y^4} | \le 1$? – Ovi Sep 17 '18 at 01:28
  • I don't think this is correct; if $0 < x, y < 1$ then the inequality in my above comment implies $0 \le x^2 + y^3(y-1)$; if $x$ is close to $0$ and $y$ is close to $1$ I think this fails. – Ovi Sep 17 '18 at 01:32
  • @Ovi, you're right. I've edited the answer. – Matheus Andrade Sep 17 '18 at 01:39