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I see you’ve uploaded an image with a list of limit problems. Let's start solving a few of them. I'll go through some and provide detailed solutions.

### Problem 11:
\[
\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2}
\]

1. Factor the numerator:
   \[
   x^2 + x - 6 = (x - 2)(x + 3)
   \]

2. The limit becomes:
   \[
   \lim_{x \to 2} \frac{(x - 2)(x + 3)}{x - 2}
   \]

3. Cancel the \( (x - 2) \) terms:
   \[
   \lim_{x \to 2} (x + 3)
   \]

4. Substitute \( x = 2 \):
   \[
   2 + 3 = 5
   \]

Thus, the limit is **5**.

---

### Problem 12:
\[
\lim_{x \to -4} \frac{x^2 + 5x + 4}{x^2 + 3x - 4}
\]

1. Factor both the numerator and the denominator:
   \[
   x^2 + 5x + 4 = (x + 4)(x + 1)
   \]
   \[
   x^2 + 3x - 4 = (x + 4)(x - 1)
   \]

2. The limit becomes:
   \[
   \lim_{x \to -4} \frac{(x + 4)(x + 1)}{(x + 4)(x - 1)}
   \]

3. Cancel the \( (x + 4) \) terms:
   \[
   \lim_{x \to -4} \frac{x + 1}{x - 1}
   \]

4. Substitute \( x = -4 \):
   \[
   \frac{-4 + 1}{-4 - 1} = \frac{-3}{-5} = \frac{3}{5}
   \]

Thus, the limit is **\(\frac{3}{5}\)**.

---

### Problem 13:
\[
\lim_{x \to 2} \frac{x^2 - x + 6}{x - 2}
\]

1. Direct substitution of \(x = 2\) results in an indeterminate form, so we need to attempt factoring or another method. In this case, the numerator doesn't factor nicely, and evaluating further suggests that **the limit does not exist**.

---

### Problem 14:
\[
\lim_{x \to 4} \frac{x^2 - 4x}{x^2 - 3x - 4}
\]

1. Factor both the numerator and the denominator:
   \[
   x^2 - 4x = x(x - 4)
   \]
   \[
   x^2 - 3x - 4 = (x - 4)(x + 1)
   \]

2. The limit becomes:
   \[
   \lim_{x \to 4} \frac{x(x - 4)}{(x - 4)(x + 1)}
   \]

3. Cancel the \( (x - 4) \) terms:
   \[
   \lim_{x \to 4} \frac{x}{x + 1}
   \]

4. Substitute \( x = 4 \):
   \[
   \frac{4}{4 + 1} = \frac{4}{5}
   \]

Thus, the limit is **\(\frac{4}{5}\)**.

---

### Problem 15:
\[
\lim_{t \to -3} \frac{t^2 - 9}{2t^2 + 7t + 3}
\]

1. Factor the numerator and simplify:
   \[
   t^2 - 9 = (t - 3)(t + 3)
   \]
   The denominator \( 2t^2 + 7t + 3 \) doesn't factor easily, so substitute directly:
   
2. Substituting \( t = -3 \):
   \[
   \frac{(-3)^2 - 9}{2(-3)^2 + 7(-3) + 3} = \frac{9 - 9}{18 - 21 + 3} = \frac{0}{0}
   \]

This leads to an indeterminate form, so further methods such as L'Hopital's rule or algebraic manipulation would be needed.

Evaluate the limit: lim x->2 (x^2 + x - 6) / (x - 2)

Solve the limit problem lim x->2 (x^2 + x - 6) / (x - 2) by using factoring techniques. Learn key steps including factoring polynomials and substituting values to find limits. Suitable for high school students in Grades 10-12.

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