https://file.aiquickdraw.com/mathbot/file/1726826790470i5zxw9zf.jpg

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.

Show that the limit lim (x,y) → (1, -1) of (xy + 1) / (x² - y²) does not exist.

Explore why the limit of the function (xy + 1) / (x² - y²) does not exist as (x, y) approaches (1, -1). Using path testing and factoring techniques, we show how different results emerge depending on the approach path. This is a detailed multivariable calculus problem suitable for undergraduate students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation