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I have to prove that

$$\lim_{(x,y) \to (0,0)} \frac{(x+y)^3}{(\sqrt{x^2-y^2})^2}$$ doesn't exist.

I think I have tried every way possible to show that doesn't exist $(x=0;y=0; x=my; x=my^2;...)$, but nothing works. I get 0 every time.

Does anyone have any suggestions? Thank you

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    There is clearly an issue when $x \approx y$ and a bigger issue when $|x| \le |y|$. Try $x=\tan y$ – 2017-02-23
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    First observe that $(\sqrt{a})^2=a$. And $$\frac{(x+y)^3}{(\sqrt{x^2-y^2})^2}=\frac{(x+y)^2}{x-y}$$ – 2017-02-23
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    @Masacroso: does $\sqrt{a}$ really exist when $a < 0$ ? – 2017-02-23
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    @Henry what about complex numbers? – 2017-02-23
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    The only thing we could say is that $(\sqrt{a})^2=|a|$ Even though I try to do that and I couldn't get to any conclusion. Ps:The problem is for real numbers – 2017-02-23
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    @Masacroso You are right, my mistake. If I consider two sets such has $x=1+y$ and $y=0 \wedge x=\mathbb{R}\backslash\{0\}$ should work, right? – 2017-02-23
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    @BA_94 I dont follow exactly what is your intention defining these sets. – 2017-02-23
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    @Masacroso I was trying to prove that for different sets, the limit is different, but on second thought I think I can't do that because on the first set if $y=0$ then $x=1$. I was trying to use the help from Fred that he deleted afterwards. – 2017-02-23
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    BA_94, you are confusing $(\sqrt a)^2$ and $\sqrt{a^2}$ (assuming we are talking about real function $x\mapsto \sqrt x$). $(\sqrt a)^2 = a$ by definition of square root, while $\sqrt{a^2} = |a|$ since square root is chosen to be nonnegative. Of course, $(\sqrt a)^2 = |a|$ is not wrong, it's just that absolute value is superfluous. – 2017-02-23

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What you can do is to find the limit along a path $y = x\gamma(x)$. Using that the expression becomes:

$${(x+y)^3\over\left(\sqrt{x^2-y^2}\right)} ={(x+x\gamma(x))^3\over\left(\sqrt{x^2-x^2\gamma(x)^2}\right)^2} = {x^3(1+\gamma(x))^3\over x^2(1-\gamma(x)^2)} = {x (1+\gamma(x))^2\over 1-\gamma(x)}$$

Now if we select $\gamma(x)$ such that $1-\gamma(x) = x/C$ (ie $\gamma(x) = 1-x/C$) we get

$${(x+y)^3\over\left(\sqrt{x^2-y^2}\right)^2}={x(1+1-x/C)^2\over 1-(1-x/C)} = {x(2-x/C)^2\over x/C}=C(2-x/C)^2 \to 4C$$

Now we have constructed paths to origin with arbitrary limit, but for the limit to exist the path-wise limit must be independent of the path we chose.

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    Thank for your help, but I didn't understand the last part. Why didn't you substitute the $k$ from the numerator? – 2017-02-23
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    @BA_94 I was a bit sloppy in my explanation and have updated the answer to be a little more elaborate. In doing so I uncovered an error, the limit should be $2C$ instead. Strictly speaking one need not substitute all $k$s since we still require that $k=1-x/C$. – 2017-02-23
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    I needed some work to check all the steps but this is correct. The exploitation of $k:=y/x$ for arbitrary $k$ is very clever. But it will be better to write $z$ instead of $k$, because you need that $k$ be a variable and the letter $k$ reminds a constant. We have that $k\to 1$. – 2017-02-23
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    @Masacroso Thanks for the feedback. My first thought about using $k$ was that I was thinking of the slope of a line, however I realize that the use of it and the association it brings is completely wrong. I don't like $z$ either as it is the third axis. I updated (improved I hope) the answer to use $\gamma(x)$ which makes it clear that it's a value that varies with $x$. – 2017-02-23