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

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.

• 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)... – Andrew Mar 17 '17 at 19:49
• Does not exist... – Lanier Freeman Mar 17 '17 at 19:55
• Your logic is sound. – Doug M Mar 17 '17 at 20:30

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.