For $f(x,y) = \frac{xy-1}{x^2 y^2-1}$, what is the limit as $(x,y)$ goes to $(1,1)$?

Question

For $f(x,y) = \frac{xy-1}{x^2 y^2-1}$, what is the limit as $(x,y)$ goes to $(1,1)$?

Since the denominator can be factored into $(xy-1)(xy+1)$ and then the $(xy-1)$'s in both the numerator and denominator can be cancelled, we are left with $f(x,y) = \frac{1}{xy+1}$ and substituting $x=1$ and $y=1$, the limit becomes $\frac{1}{2}$. Is this solution right? We had this question on our test todayin Multivariable Caclulus, and some people put $1/2$ while others put $DNE$. Who is right?

Answer 1

Yes it is right, indeed we have that as $(x,y) \to (1,1)$ also $t=xy \to 1$ therefore we reduce to a limit for a single variable that is

$$\frac{xy-1}{x^2 y^2-1}=\frac{t-1}{t^2-1}=\frac{t-1}{(t+1)(t-1)}=\frac{1}{t+1}\to \frac12$$

As noticed in the comments $f(x,y)$ fails to exist along the hyperbolas $y=\pm\frac1x$ but it doesn't matter since these points are not in the domain.

Answer 2

This is right. Indeed the limit exists and it is 1/2.

Your line of reasoning is about right. A critical observation needed (which I think you used implicitly) is that for any $\epsilon$, there is a $\delta$ s.t. long as $||(x,y)-(1,1)|| \le \delta$, then $|1-\frac{1}{xy+1}| \le \epsilon$.

Answer 3

Consider $X=x-1$ and $Y=y-1$, then we have $$ \begin{array}{c} \lim_{(x,y)\to (1,1)}\frac{xy-1}{x^2 y^2-1}= \lim_{(X,Y)\to (0,0)}\frac{XY-Y-X}{ \left( XY-Y-X+2 \right) \left( XY-Y-X \right)} =\lim_{(X,Y)\to (0,0)}\frac{1}{ \left( XY-Y-X+2 \right)}=\frac{1}{2} \end{array} $$