We are tasked with finding the limit of the function $\frac{x + 1}{1 - x^2}$ as $x$ approaches $-1$.

Step 1: Write the expression

The given expression is: $\lim_{x \to -1} \frac{x + 1}{1 - x^2}$

Step 2: Simplify the denominator

Notice that $1 - x^2$ is a difference of squares: $1 - x^2 = (1 - x)(1 + x)$

Thus, the expression becomes: $\lim_{x \to -1} \frac{x + 1}{(1 - x)(1 + x)}$

Step 3: Cancel common factors

We can cancel out the common factor of $(x + 1)$ in the numerator and denominator: $\lim_{x \to -1} \frac{1}{1 - x}$

Step 4: Substitute $x = -1$ into the simplified expression

Now, substitute $x = -1$ into the simplified expression: $\frac{1}{1 - (-1)} = \frac{1}{1 + 1} = \frac{1}{2}$

Conclusion:

The limit is: $\boxed{\frac{1}{2}}$

Would you like more details, or do you have any further questions?

Here are 5 related questions for practice:

1. Find the limit as $x \to 1$ of $\frac{x - 1}{1 - x^2}$.
2. Evaluate the limit $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$.
3. Determine the limit $\lim_{x \to 0} \frac{x^2 + 2x}{x}$.
4. Compute $\lim_{x \to -1} \frac{x^2 - 1}{x + 1}$.
5. Find the limit ( \lim_{x \to 3} \frac{x^2 - 9}{x -