The Question asks: Find the delta epsilon limit if it exist $$ \lim \limits_{(x, y) \to (0, 0)} \ \frac{x^2 y^3}{2x^2 + y^2} = 1$$

What I've done so far:

$$ \mid {\frac{x^2 y^3}{2x^2 + y^2} - 1 \mid} < \epsilon \quad \quad \quad \quad0 < \sqrt{x^2 + y^2} < \delta $$

Now, since $x ^2 \leq 2x^2 + y^ \implies \frac{x^2}{2x^2 + y^2} \leq 1$ Then: $$ \frac{x^2 \mid y^3 \mid}{2x^2 + y^2} \leq \mid y^3 \mid = \sqrt{y^2}^3 = \quad ...$$

Here is where I am stuck, I can't make $\sqrt{y^2}^3$ into $\sqrt{ x^2 + y^2} $.

How should I proceed further?

Using Stewart's Calculus (8th ed) 14.2 Limit proof method


2 Answers 2


For both $x,y\ne 0$, then \begin{align*} \dfrac{|x|^{2}|y|^{3}}{2x^{2}+y^{2}}&\leq\dfrac{|x|^{2}|y|^{3}}{2\sqrt{2}|x||y|}\\ &=\dfrac{1}{2\sqrt{2}}|x||y|^{2}\\ &\leq\dfrac{1}{2\sqrt{2}}\sqrt{|x|^{2}+|y|^{2}}(|x|^{2}+|y|^{2}), \end{align*} if either $x=0$ or $y=0$ (but not both), the above inequality is still true. So \begin{align*} \dfrac{|x|^{2}|y|^{3}}{2x^{2}+y^{2}}\leq\dfrac{1}{2\sqrt{2}}\sqrt{|x|^{2}+|y|^{2}}(|x|^{2}+|y|^{2}),~~~~(x,y)\ne(0,0). \end{align*} Given $\epsilon>0$, for all $0<\sqrt{|x|^{2}+|y|^{2}}<2^{1/2}\epsilon^{1/3}$, then $\dfrac{|x|^{2}|y|^{3}}{2x^{2}+y^{2}}\leq\dfrac{1}{2\sqrt{2}}\sqrt{|x|^{2}+|y|^{2}}(|x|^{2}+|y|^{2})=\dfrac{1}{2\sqrt{2}}(|x|^{2}+|y|^{2})^{3/2}<\epsilon$.

The limit is zero.

Another way: \begin{align*} \dfrac{x^{2}}{2x^{2}+y^{2}}\leq 1,~~~~(x,y)\ne(0,0), \end{align*} so \begin{align*} \dfrac{x^{2}|y|^{3}}{2x^{2}+y^{2}}\leq|y|^{3}=(y^{2})^{3/2}\leq(x^{2}+y^{2})^{3/2}. \end{align*}

  • $\begingroup$ I'm sorry I don't understand this proof method; How did you get $\frac{1}{2\sqrt{2}}$ as the value? Shouldn't it be $\frac{1}{\sqrt{2}}$ Then subsequently, how did you move from step 2 to 3? I think that has to do with the triangle inequality (?) thank you $\endgroup$
    – Andre Fu
    Mar 25, 2018 at 1:46
  • $\begingroup$ The first inequality is Arithmetic-Geometric which asserts that $a+b\geq 2\sqrt{ab}$. The second inequality follows by $|x|^{2}\leq|x|^{2}+|y|^{2}$, so $|x|\leq\sqrt{|x|^{2}+|y|^{2}}$. Indeed, $|y|^{2}\leq|x|^{2}+|y|^{2}$ is another one. $\endgroup$
    – user284331
    Mar 25, 2018 at 1:50
  • $\begingroup$ Your second way is the method that I was using, how does one relate the $ (x^2 + y^2)^{3/2} $ to $ \epsilon $ $\endgroup$
    – Andre Fu
    Mar 25, 2018 at 18:07
  • $\begingroup$ With $x^{2}+y^{2}<\epsilon^{2/3}$, then $(x^{2}+y^{2})^{3/2}<\epsilon$, done. $\endgroup$
    – user284331
    Mar 25, 2018 at 18:12
  • $\begingroup$ I mean, let $\delta=\epsilon^{1/3}$, for $\sqrt{x^{2}+y^{2}}<\delta=\epsilon^{1/3}$, then $x^{2}+y^{2}<\epsilon^{2/3}$... $\endgroup$
    – user284331
    Mar 25, 2018 at 18:13

In fact

$$\left|\frac{x^2y^3}{2x^2+y^2}\right| \leq \frac{1}{2}|y^3|$$ which is equivalent to

$$2x^2\leq 2x^2+y^2.$$

  • $\begingroup$ This is very helpful, thank you. This still doesn't solve the problem of $$ \frac{1}{2} \mid y^3 \mid = \frac{1}{2}(\sqrt{y^2})^3 $$ How does one proceed from there? $\endgroup$
    – Andre Fu
    Mar 25, 2018 at 5:40
  • $\begingroup$ I dont understand the problem. And $\epsilon-\delta$ proof is clear from the formula. $\endgroup$ Mar 25, 2018 at 12:25
  • $\begingroup$ A hate downvote... $\endgroup$ Apr 3, 2018 at 20:11
  • $\begingroup$ I'm sorry, but I was not the one to downvote! I have accepted user284331's solution since last week! $\endgroup$
    – Andre Fu
    Apr 3, 2018 at 20:23
  • $\begingroup$ No, I never thought so. I downvoted another question, and gave my reasons, then suddenly two of my questions from last week, and on different subjects are downvoted in the same minute. $\endgroup$ Apr 3, 2018 at 20:30

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