# Multivariable delta-epsilon limit proofs

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

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*}

• 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 Mar 25, 2018 at 1:46
• 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. Mar 25, 2018 at 1:50
• Your second way is the method that I was using, how does one relate the $(x^2 + y^2)^{3/2}$ to $\epsilon$ Mar 25, 2018 at 18:07
• With $x^{2}+y^{2}<\epsilon^{2/3}$, then $(x^{2}+y^{2})^{3/2}<\epsilon$, done. Mar 25, 2018 at 18:12
• I mean, let $\delta=\epsilon^{1/3}$, for $\sqrt{x^{2}+y^{2}}<\delta=\epsilon^{1/3}$, then $x^{2}+y^{2}<\epsilon^{2/3}$... 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

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

• 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? Mar 25, 2018 at 5:40
• I dont understand the problem. And $\epsilon-\delta$ proof is clear from the formula. Mar 25, 2018 at 12:25
• A hate downvote... Apr 3, 2018 at 20:11
• I'm sorry, but I was not the one to downvote! I have accepted user284331's solution since last week! Apr 3, 2018 at 20:23
• 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. Apr 3, 2018 at 20:30