Let's go through each of the limits one by one.

1)

$\lim_{x \to -1} \frac{x^2 + 2x - 3}{x - 1}$

First, substitute $x = -1$ into the expression: $\frac{(-1)^2 + 2(-1) - 3}{-1 - 1} = \frac{1 - 2 - 3}{-2} = \frac{-4}{-2} = 2.$ So, the limit is: $\boxed{2}.$

2)

$\lim_{x \to 1} \frac{x^2 + 2x - 3}{x - 1}$

Substitute $x = 1$ into the expression: $\frac{1^2 + 2(1) - 3}{1 - 1} = \frac{1 + 2 - 3}{0} = \frac{0}{0}.$ This is an indeterminate form, so we need to simplify the expression. Factor the numerator: $x^2 + 2x - 3 = (x - 1)(x + 3).$ Thus, the limit becomes: $\lim_{x \to 1} \frac{(x - 1)(x + 3)}{x - 1}.$ Cancel out $(x - 1)$ from the numerator and denominator: $\lim_{x \to 1} (x + 3) = 1 + 3 = 4.$ So, the limit is: $\boxed{4}.$

3)

$\lim_{x \to 1} \frac{x^2 - x - 1}{x - 1}$

Substitute $x = 1$ into the expression: $\frac{1^2 - 1 - 1}{1 - 1} = \frac{1 - 1 - 1}{0} = \frac{-1}{0}.$ This is an indeterminate form, and division by zero suggests that the limit might be infinite. To investigate the behavior, let's factor the numerator: $x^2 - x - 1 = (x - 1)(x + 1).$ Thus, the limit becomes: $\lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1}.$ Cancel out $(x - 1)$ from the numerator and denominator: $\lim_{x \to 1} (x + 1) = 1 + 1 = 2.$ So, the limit is: $\boxed{2}.$

4)

$\lim_{x \to \infty} \frac{2x^3 + 3x - 5}{-2x^2 + 2x + 7}$

For large values of $x$, the highest-degree terms dominate both the numerator and denominator. In the numerator, the highest-degree term is $2x^3$, and in the denominator, it is $-2x^2$. To simplify the limit, we divide both the numerator and denominator by $x^2$, the highest power of $x$ in the denominator:

$\frac{2x^3 + 3x - 5}{-2x^2 + 2x + 7} = \frac{2x + \frac{3}{x} - \frac{5}{x^2}}{-2 + \frac{2}{x} + \frac{7}{x^2}}.$

As $x \to \infty$, the terms $\frac{3}{x}$, $\frac{5}{x^2}$, $\frac{2}{x}$, and $\frac{7}{x^2}$ all approach 0. So, the limit simplifies to:

$\lim_{x \to \infty} \frac{2x}{-2} = -x.$

As $x \to \infty$, $-x$ approaches $-\infty$. Therefore, the limit is: $\boxed{-\infty}.$

---

Final Answers:

1. $\lim_{x \to -1} \frac{x^2 + 2x - 3}{x - 1} = 2$
2. $\lim_{x \to 1} \frac{x^2 + 2x - 3}{x - 1} = 4$
3. $\lim_{x \to 1} \frac{x^2 - x - 1}{x - 1} = 2$
4. $\lim_{x \to \infty} \frac{2x^3 + 3x - 5}{-2x^2 + 2x + 7} = -\infty$

Would you like further clarification or details on any of these steps?

---

Here are some related questions:

1. How would you find the limit of a rational function when it results in a $\frac{0}{0}$ indeterminate form?
2. What is the significance of dividing by the highest power of $x$ when finding limits at infinity?
3. How can you determine the behavior of a rational function near vertical asymptotes?
4. How do you interpret limits that approach positive or negative infinity?
5. What is the role of polynomial factorization in simplifying rational expressions for limits?

Tip: Always factor out common terms and check for simplifications before evaluating limits that lead to indeterminate forms like $\frac{0}{0}$.