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It looks like you've uploaded an image containing various limit problems. Since parts of the text are blocked, I'll work with the visible problems to help you solve them. Let's start by solving some of the visible limits.

### Problem 2:
\[
\lim_{{x \to 1}} \frac{{x - 1}}{{x^2 + 2x - 3}}
\]

1. Factor the denominator: \(x^2 + 2x - 3 = (x - 1)(x + 3)\).
2. The expression becomes:
\[
\lim_{{x \to 1}} \frac{{x - 1}}{{(x - 1)(x + 3)}}
\]
3. Cancel the \((x - 1)\) terms:
\[
\lim_{{x \to 1}} \frac{1}{{x + 3}}
\]
4. Substituting \(x = 1\), we get:
\[
\frac{1}{1 + 3} = \frac{1}{4}
\]

So, the answer is \(\frac{1}{4}\).

---

### Problem 4:
\[
\lim_{{x \to 1}} \frac{{x^2 - 3x + 2}}{{4x + 3}}
\]

1. Factor the numerator: \(x^2 - 3x + 2 = (x - 1)(x - 2)\).
2. The limit becomes:
\[
\lim_{{x \to 1}} \frac{{(x - 1)(x - 2)}}{{4x + 3}}
\]
3. Substituting \(x = 1\) directly into the simplified expression:
\[
\frac{{(1 - 1)(1 - 2)}}{{4(1) + 3}} = \frac{{0 \times (-1)}}{{7}} = 0
\]

So, the answer is \(0\).

---

### Problem 8:
\[
\lim_{{x \to 3}} \frac{{\sqrt{2x + 3} - 3}}{{x - 3}}
\]

1. Apply the conjugate multiplication technique. Multiply the numerator and denominator by the conjugate of the numerator:
\[
\frac{{\sqrt{2x + 3} - 3}}{{x - 3}} \times \frac{{\sqrt{2x + 3} + 3}}{{\sqrt{2x + 3} + 3}}
\]
2. Simplifying the numerator:
\[
(\sqrt{2x + 3})^2 - 3^2 = (2x + 3) - 9 = 2x - 6
\]
3. So, the expression becomes:
\[
\frac{{2(x - 3)}}{{(x - 3)(\sqrt{2x + 3} + 3)}}
\]
4. Cancel the \((x - 3)\) terms:
\[
\lim_{{x \to 3}} \frac{2}{{\sqrt{2x + 3} + 3}}
\]
5. Substituting \(x = 3\):
\[
\frac{2}{{\sqrt{2(3) + 3} + 3}} = \frac{2}{{\sqrt{6 + 3} + 3}} = \frac{2}{{3 + 3}} = \frac{2}{6} = \frac{1}{3}
\]

So, the answer is \(\frac{1}{3}\).

---

These are the solutions to the visible limits. Would you like a deeper explanation or help with more of the problems?

Here are 5 related questions you could explore:
1. How do you factor polynomials to simplify limit expressions?
2. What other techniques can be used to solve indeterminate limits of the form \(\frac{0}{0}\)?
3. How does L'Hôpital's Rule apply to limits involving indeterminate forms?
4. How do you apply conjugate multiplication to limits involving square roots?
5. What strategies are effective when solving limits involving rational expressions?

**Tip**: Always check if direct substitution gives an indeterminate form before applying further techniques.

Step-by-step solutions to limit problems such as \lim_{{x \to 1}} \frac{{x - 1}}{{x^2 + 2x - 3}} and \lim_{{x \to 3}} \frac{{\sqrt{2x + 3} - 3}}{{x - 3}}, including techniques like polynomial factorization and conjugate multiplication. Ideal for high school students in grades 10-12.

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