https://file.aiquickdraw.com/mathbot/file/1725581279475ziml333e.jpg

The problem presented in the image is:

\[
\lim_{x \to 3} \frac{6}{x^2 + 9}
\]

Let's solve it step by step:

### Step 1: Substituting \(x = 3\) into the expression
We substitute \(x = 3\) directly into the expression since there is no indeterminate form (like \(\frac{0}{0}\)):

\[
\frac{6}{(3)^2 + 9} = \frac{6}{9 + 9} = \frac{6}{18}
\]

### Step 2: Simplifying the expression

\[
\frac{6}{18} = \frac{1}{3}
\]

### Final Answer:
\[
\lim_{x \to 3} \frac{6}{x^2 + 9} = \frac{1}{3}
\]

Let me know if you want further details or clarifications!

Here are 5 related questions to expand your understanding:
1. What happens to the limit when \(x \to -3\) for the same function?
2. How do limits behave when the numerator approaches zero while the denominator is a non-zero constant?
3. What if the limit approached \(x \to 0\) instead of 3 for this same function?
4. How can you find the limit for a rational function when both numerator and denominator involve polynomials?
5. How does the concept of continuity relate to limits?

**Tip:** Always check if direct substitution leads to a defined value before attempting more complex limit techniques.

Solve the limit lim(x → 3) 6 / (x² + 9) using direct substitution. This problem demonstrates the application of basic limit theorems and algebraic manipulation. Suitable for high school students in Grades 10-12.

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