To solve the limit of $\frac{x-1}{x+1}$ as $x$ approaches $-1$, we follow these steps:

1. Direct Substitution: Substitute $x = -1$ into the expression:

   $\frac{-1 - 1}{-1 + 1} = \frac{-2}{0}$

   This results in a division by zero, indicating that we need further analysis.

2. Behavior Analysis: Analyze the behavior of the function as $x$ approaches $-1$ from both the left and the right sides.

   • As $x$ approaches $-1$ from the left ($x \to -1^-$): The numerator $(x-1)$ approaches $-2$, and the denominator $(x+1)$ approaches $0$ from the negative side. Thus, $\frac{x-1}{x+1}$ approaches $+\infty$.

   • As $x$ approaches $-1$ from the right ($x \to -1^+$): The numerator $(x-1)$ approaches $-2$, and the denominator $(x+1)$ approaches $0$ from the positive side. Thus, $\frac{x-1}{x+1}$ approaches $-\infty$.

3. Conclusion: Since the limits from the left and the right are not equal, the limit does not exist.

Thus, the limit of $\frac{x-1}{x+1}$ as $x$ approaches $-1$ does not exist.

Would you like more details or have any questions?

Here are some more questions you might be interested in:

1. How to evaluate the limit of a function as $x$ approaches infinity?
2. How to find the derivative of a given function?
3. How to integrate a given function?
4. What is the Taylor series expansion of a function?
5. How to solve a differential equation?

Tip: When evaluating limits that lead to indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, consider using L'Hôpital's Rule.