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Given is the function $f(x,y)=\frac{x^3-y^3}{x^2-y^2}$ defined for $(x,y)\in\mathbb{R^2}, \, x^2\neq y^2$

I need to find the limit of $\lim_{(x,y)\rightarrow(0,0)} \frac{x^3-y^3}{x^2-y^2}$.

I have tried a few paths, $(x,mx),\, (x,mx^2), \, (x,sin(x)), \, (x,tan(x))$ but all of them yield the same result, $0$.

So I tried to prove the limit exists using the epsilon-delta definition. So: $$\left|\frac{x^3-y^3}{x^2-y^2} - 0\right| = \left|\frac{(x-y)(x^2+y^2+xy)}{(x-y)(x+y)}\right|= \frac{x^2+y^2+xy}{\left|x+y\right|}<\epsilon$$ Here I am stuck and do not know how to proceed.

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    $\begingroup$ on (x,-x) is not even defined.. $\endgroup$ – Exodd Feb 20 '17 at 10:23
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The problem is that there is no limit, only pathwise limits (which of course depend on the path to zero).

To see this one could rewrite the expression:

$${x^3-y^3\over x^2-y^2} = {(x-y)(x^2+xy+y^2)\over(x-y)(x+y)} = {(x+y)x - y^2\over (x+y)} = x - {y^2\over x+y}$$

Now of course $x\to 0$, but the term $y^2/(x+y)$ can be selected arbitrarily by having $x = y^2/C - y$. So along the parabola $x = y^2/C - y$ aproaching $0$ we have the limit of the expression being $-C$.

But this can't happen if the limit actually exists.

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  • $\begingroup$ Shouldn't the limit of the expression equal $-C$? $\endgroup$ – PaRaXeRoX Feb 20 '17 at 11:50
  • $\begingroup$ @PaRaXeRoX You're correct - I've corrected that... $\endgroup$ – skyking Feb 20 '17 at 12:10
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A method is to consider $y=mx$

Note : $m=m(x,y)$ is variable, it is the ratio of converging rates of $y$ and $x$.


$$f(x,y)=\frac{x^2(1+m^2+m)}{x(1+m)}=x\frac{1+m+m^2}{1+m}$$

If $m\to\pm\infty\quad f(x,y)\sim\frac{m^2}{m}x=mx=y\to 0$

If $m\to0$ then $f(x,y)\sim x\to 0$

If $m$ is bounded let's have $u=m+1\quad$ ($u$ also variable).

$\displaystyle{f(x,y)=x\frac{u+(u-1)^2}{u}=x\frac{u^2-u+1}{u}=x(u-1+\frac1u)=xm+\frac xu\sim \frac xu}$

because if $m$ is bounded then $mx\to 0$.

And that is a big problem : for $u=kx$ with $k\in\mathbb R$ we have $f(x,y)\to\frac1k$ and $m=kx-1$ is effectively bounded, thus a valid case.

Conclusion :

There is no limit in $(0,0)$ because for $k\in\mathbb R,\ x\to 0,\ f(x,kx^2-x)\to\frac1k$

$f$ has multiple limits depending on the path chosen to go to $(0,0)$.

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