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The given problem asks to find the constant \( C \) such that the limit of the following expression exists:

\[
\lim_{x \to 3} \frac{x^2 + x + C}{x^2 - 5x + 6}
\]

### Step 1: Factor the Denominator
The denominator is \( x^2 - 5x + 6 \). We can factor this as:

\[
x^2 - 5x + 6 = (x - 2)(x - 3)
\]

So the expression becomes:

\[
\frac{x^2 + x + C}{(x - 2)(x - 3)}
\]

### Step 2: Analyze the Limit
As \( x \to 3 \), the denominator \( (x - 3) \) goes to zero. For the limit to exist, the numerator must also go to zero at \( x = 3 \) to cancel the zero in the denominator.

### Step 3: Set the Numerator Equal to Zero at \( x = 3 \)
Substitute \( x = 3 \) into the numerator \( x^2 + x + C \):

\[
3^2 + 3 + C = 0
\]

This simplifies to:

\[
9 + 3 + C = 0 \quad \Rightarrow \quad 12 + C = 0 \quad \Rightarrow \quad C = -12
\]

### Final Answer:
The constant \( C \) that ensures the limit exists is \( C = -12 \).

Would you like a more detailed explanation on any specific part, or do you have additional questions?

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Here are five related questions for further exploration:

1. How do we handle indeterminate forms in limits?
2. What happens if we don't factor the denominator correctly?
3. Why does the numerator need to be zero at \( x = 3 \) for the limit to exist?
4. What is the importance of factoring in rational expressions when solving limits?
5. How can we apply L'Hopital's Rule if the limit is indeterminate?

**Tip:** Always check if direct substitution leads to an indeterminate form when solving limits before applying any factorization or algebraic manipulation.

Encuentre la constante C tal que lim (x -> 3) (x^2 + x + C) / (x^2 - 5x + 6) exista.

In this problem, we find the constant C that ensures the limit of the rational expression (x^2 + x + C) / (x^2 - 5x + 6) exists as x approaches 3. We explore concepts of limits, factoring, and how to ensure limits exist by canceling terms. Suitable for Grades 10-12.

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