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Let's solve each of these limits step by step.

---

### Problem (a)

\[
\lim_{{x \to \frac{1}{3}}} \frac{9x^2 - 1}{3x + 1}
\]

1. **Substitute \( x = \frac{1}{3} \):**

   \[
   9\left(\frac{1}{3}\right)^2 - 1 = 9 \cdot \frac{1}{9} - 1 = 1 - 1 = 0
   \]
   
   and

   \[
   3\left(\frac{1}{3}\right) + 1 = 1 + 1 = 2.
   \]

   Substituting directly does not yield an indeterminate form since we only get 0 in the numerator and not in the denominator. Therefore, the limit is

   \[
   \frac{0}{2} = 0.
   \]

So, the answer to (a) is:
\[
0.
\]

---

### Problem (b)

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

1. **Substitute \( x = 1 \):**

   \[
   x^3 + 1 = 1^3 + 1 = 2
   \]
   
   and

   \[
   x^2 - 1 = 1^2 - 1 = 0.
   \]

   This results in an indeterminate form \( \frac{2}{0} \), so we need to simplify further. Factoring the denominator, we have

   \[
   x^2 - 1 = (x - 1)(x + 1).
   \]

   The expression becomes:

   \[
   \frac{x^3 + 1}{(x - 1)(x + 1)}.
   \]

To continue from here,

Calculate the following limits: 
(a) \( \lim_{x \to \frac{1}{3}} \frac{9x^2 - 1}{3x + 1} \) 
(b) \( \lim_{x \to 1} \frac{x^3 + 1}{x^2 - 1} \) 
(c) \( \lim_{x \to 2} \left( \frac{1}{x} - \frac{1}{2} \right) \div (x - 2) \)

Learn how to solve limit problems involving algebraic expressions using direct substitution and factoring techniques. The problems cover limits where substitution initially leads to indeterminate forms, requiring further simplification. Suitable for high school students in Grades 11-12.

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