The problem asks to show that the limit

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

does not exist.

Step-by-Step Approach:

1. Expression Simplification:

We start by simplifying the given expression. Notice that the denominator $x^2 - y^2$ is a difference of squares, so we can factor it:

$x^2 - y^2 = (x - y)(x + y)$

Thus, the expression becomes:

$\frac{xy + 1}{(x - y)(x + y)}$

2. Testing Along Different Paths:

To show that the limit does not exist, we can evaluate the limit along different paths approaching the point $(1, -1)$ and see if we get different results.

• Path 1: Along $y = -x$:

  Substitute $y = -x$ into the expression:

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

  As $x \to 1$, the denominator tends to 0 while the numerator tends to $1 - 1 = 0$, creating an indeterminate form. This suggests that along this path, the expression does not yield a valid limit.

• Path 2: Along $x = 1$:

  Now, substitute $x = 1$ into the expression:

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

  As $y \to -1$, the numerator $y + 1 \to 0$ and the denominator $1 - y^2 \to 0$, leading to an indeterminate form. Applying L'Hopital's Rule here could yield further insights, but at this stage, we can already see divergence.

• Path 3: Along $y = -1$:

  If we substitute $y = -1$:

  $\frac{xy + 1}{(x - y)(x + y)} = \frac{x(-1) + 1}{(x + 1)(x - 1)} = \frac{-x + 1}{(x + 1)(x - 1)}$

  As $x \to 1$, the numerator tends to 0, but the denominator also tends to 0. Again, this leads to an indeterminate form, which suggests the possibility of no limit.

3. Conclusion:

Since evaluating along different paths gives inconsistent or undefined results, the limit does not exist.

Would you like more details on any of these paths, or have any questions?

Here are five related questions to deepen your understanding:

1. What are the general conditions under which a two-variable limit does not exist?
2. How can the path method help determine the existence of a limit?
3. What are other techniques besides the path method for proving a limit does not exist?
4. How would this problem change if the expression had a different form in the denominator?
5. Can the use of polar coordinates simplify solving two-variable limits?

Tip: When proving that a limit does not exist, testing multiple paths is a powerful approach to showing divergence.