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.