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The limit problem you provided is:

\[
\lim_{{x \to 3}} \frac{2x^2 + 4x - 5}{x^2 - 5x + 6}
\]

### Step-by-Step Solution:

1. **Check for direct substitution**:
   Substitute \( x = 3 \) directly into the expression.

   \[
   \text{Numerator: } 2(3)^2 + 4(3) - 5 = 18 + 12 - 5 = 25
   \]
   \[
   \text{Denominator: } (3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0
   \]

   Since the denominator is 0 and the numerator is 25, this results in a form of \(\frac{25}{0}\), indicating that the limit does not exist in the conventional sense. However, we can explore whether it tends to \( \infty \) or \( -\infty \) by analyzing the behavior near \( x = 3 \).

2. **Factor the denominator**:
   Factor the quadratic expression in the denominator.

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

3. **Analyze the behavior near \( x = 3 \)**:
   We now have:

   \[
   \frac{2x^2 + 4x - 5}{(x - 2)(x - 3)}
   \]

   As \( x \to 3 \), the factor \( (x - 3) \) in the denominator approaches 0, which causes the whole expression to blow up. To determine the sign of the limit, we examine the behavior from the left (\( x \to 3^- \)) and from the right (\( x \to 3^+ \)).

   - **As \( x \to 3^- \)** (approaching from the left), \( (x - 3) \) is negative, and the rest of the expression remains positive, so the limit tends to \( -\infty \).
   - **As \( x \to 3^+ \)** (approaching from the right), \( (x - 3) \) is positive, and the rest of the expression remains positive, so the limit tends to \( +\infty \).

4. **Conclusion**:
   Since the limit approaches \( -\infty \) from the left and \( +\infty \) from the right, the limit does not exist.

### Final Answer:
\[
\lim_{{x \to 3}} \frac{2x^2 + 4x - 5}{x^2 - 5x + 6} = \text{does not exist (DNE)}
\]

Would you like more details on this or have any other questions? Here are some related questions for further exploration:

1. How do you determine whether a limit approaches infinity or negative infinity?
2. What is the significance of factoring when solving limits involving rational functions?
3. What does it mean when a limit does not exist in terms of real-world applications?
4. How can L'Hôpital's Rule be applied in limit problems with indeterminate forms?
5. What are the conditions for applying L'Hôpital's Rule?

**Tip:** When solving limits with rational functions, always try direct substitution first to identify potential indeterminate forms. Then factor the expression to simplify further.

lim (x -> 3) (2x^2 + 4x - 5) / (x^2 - 5x + 6)

Learn how to solve the limit problem lim (x -> 3) (2x^2 + 4x - 5) / (x^2 - 5x + 6) with a detailed step-by-step explanation. This includes factoring, direct substitution, and analysis of behavior around the singularity point. Suitable for Grades 11-12.

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