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To solve the limit \(\lim_{x \to 2} \frac{3x^2 + 1}{x^2 + 5}\), we can directly substitute \(x = 2\) into the function since this limit doesn't lead to an indeterminate form.

1. **Substitute \(x = 2\) into the numerator and denominator:**

   \[
   \text{Numerator: } 3x^2 + 1 = 3(2)^2 + 1 = 3(4) + 1 = 12 + 1 = 13
   \]

   \[
   \text{Denominator: } x^2 + 5 = (2)^2 + 5 = 4 + 5 = 9
   \]

2. **Calculate the limit:**

   \[
   \lim_{x \to 2} \frac{3x^2 + 1}{x^2 + 5} = \frac{13}{9}
   \]

The limit is \(\frac{13}{9}\).

Would you like further details or have any questions? Here are some related questions that might interest you:

1. How do you solve limits involving indeterminate forms like \(\frac{0}{0}\)?
2. What are the basic rules for computing limits?
3. How can L'Hôpital's Rule be applied to solve limits?
4. What are one-sided limits and how do they differ from two-sided limits?
5. How do you evaluate limits at infinity?

**Tip:** When dealing with limits, always check if direct substitution is possible before applying more advanced techniques like factoring or L'Hôpital's Rule.

Learn how to solve the limit problem \(\lim_{x \to 2} \frac{3x^2 + 1}{x^2 + 5}\) with a detailed step-by-step solution. This problem involves direct substitution and covers key concepts in limits. Suitable for students in Grades 11-12.

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