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I need to prove that: $$ \lim_{(x,y) \to (0,0)} \frac{3x^2y^2}{x^4+y^4}=\frac{3}{2} $$ using the $\epsilon$-$\delta$ notation.

I have tried everything I could think of to make the expression into a function of $x^2+y^2$ so that I could then calculate $\delta$ in terms of $\epsilon$.

Any help would be greatly appreciated!

P.S: $f(x,y)=0$ at $(x,y)=(0,0)$

Answer: This limit does not exist! If we calculate the limit along the curves $y=x$ and $y=x^2$, we get different values for the limit.

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  • $\begingroup$ What is the value of the function at $(0, 0)$? $\endgroup$ – Zhanxiong Jul 21 '16 at 15:07
  • $\begingroup$ It is not relevant. $\endgroup$ – smcc Jul 21 '16 at 15:08
  • $\begingroup$ Consider the limit along different curves, e.g. $y=ax$, $y=x^2$, $y=0$, etc. Do they always give the same limit? $\endgroup$ – jdods Jul 21 '16 at 15:14
  • $\begingroup$ The limit $$ \lim_{(x,y) \to (0,0)} \frac{3xy}{x^2+y^2}=0 $$ is well known to be path dependent, and is a mere substitution away from your limit $\endgroup$ – Brevan Ellefsen Jul 21 '16 at 15:15
  • $\begingroup$ @jdods Thanks, I don't know why I didn't test this to see if the limit exists first! $\endgroup$ – Ali Jul 21 '16 at 15:20
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To show the statement is false, notice that for all $x\neq 0$: $f(x,x)=3/2$ but $f(x,2x)=12/17$, so the limit does not exist.

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Hint: You are trying to prove something that is false.

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  • $\begingroup$ oops! Thanks alot. I don't know why I didn't notice that! $\endgroup$ – Ali Jul 21 '16 at 15:13
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Another approach that is sometimes helpful: use polar coordinates

$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\implies\frac{3x^2y^2}{x^4+y^4}=\frac{3\cos^2t\sin^2t}{\underbrace{\cos^4t+\sin^4t}_{=(\cos^2t+\sin^2t)^2-2\cos^2t\sin^2t}}=$$

$$=\frac{\frac34\sin^22t}{1-\frac12\sin^22t}$$

Either way, it is clear that $\;r\to0\implies\;$ the limit depends on the angle $\;t\;$, and thus the wanted limit doesn't exist.

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take the way $y=x$, then the limit become: $$\lim_{x\to 0}\frac{3x^2x^2}{x^4+x^4}=\lim_{x\to 0}\frac{3x^4}{2x^4}=\lim_{x\to 0}\frac{3}{2}=\frac{3}{2}$$ clearly the limit is not $0$

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  • $\begingroup$ Shouldn't it be $x^4 + x^4$ ? Not $x^2 + x^2$ ? Or maybe this is what the OP meant? Because it works out to be what the OP wanted $\endgroup$ – Joshua Lochner Jul 21 '16 at 19:08
  • $\begingroup$ @JoshuaLochner thanks a lot, i fix the problem $\endgroup$ – julio godoy Jul 23 '16 at 2:35

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