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I was asked to solve the following bivariate limit. I'd like to know if my approach is valid and if there is a better way to do it

$$\lim_{(x,y)\to(1,0)}\frac{xy-y}{x^2-2x+1+y^2}$$

I used a phase shift and rewrote as this:

$$\lim_{(x,y)\to(0,0)}\frac{(x+1)y-y}{(x+1)^2-2(x+1)+1+y^2}$$

Simplify a little:

$$\lim_{(x,y)\to(0,0)}\frac{xy+y-y}{x^2+2x+1-2x-2+1+y^2}$$

$$\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+y^2}$$

Convert to polar:

$$\lim_{r\to 0}\frac{r^2\cdot\cos\theta\cdot\sin\theta}{r^2}$$

$$\lim_{r\to0}\cos\theta\cdot\sin\theta$$

Since our answer is in terms of $\theta$, the limit DNE.

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    $\begingroup$ I think it's fine. Another way is e.g. to note that along the $x$ axis, the limit is $0$, while along the line $x=y$, the limit is $\lim_{x\to 0} \frac{x^2}{2x^2} = \frac{1}{2}$. So as you said, the limit does not exist. (Both of these are after your shift and simplification)... $\endgroup$ – Andrew Mar 17 '17 at 19:49
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    $\begingroup$ Does not exist... $\endgroup$ – Lanier Freeman Mar 17 '17 at 19:55
  • $\begingroup$ Your logic is sound. $\endgroup$ – Doug M Mar 17 '17 at 20:30
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Since you asked for a better way to do it, I'll offer a way to see it that's more direct, although I really do like your approach better than this one, since your approach does not involve looking at particular paths. We'll chalk it up to subjectivity though.

After making the change of variables and simplifying, you have the limit $\lim_{(x,y) \to (0,0)} \frac{xy}{x^2 + y^2}$. Along the line $y = 0$, this reduces to the limit $\lim_{x \to 0} \frac{0}{x^2} = 0$. On the other hand, along the line $y = x$we have $\lim_{x \to 0} \frac{x^2}{2x^2} = 1/2$. This difference is direction violates the definition of the limit, as in any small neighborhood of $0$, the values get arbitrarily close to both $0$ and $1/2$.

This is usually how people prove non-existence. Choosing $y$ to be some polynomial in $x$, or choosing one of them to be $0$ so that the limit is easy to evaluate along that line/curve, and trying to force the limit to take two different values at the same point as we did here.

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