To find the limit of the given rational function:

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

Step 1: Evaluate the direct substitution

We first check what happens if we directly substitute $x = 3$ into the function:

• For the numerator: $2(3)^2 + 4(3) - 5 = 2(9) + 12 - 5 = 18 + 12 - 5 = 25$
• For the denominator: $(3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$

Since the denominator is zero, we have an indeterminate form $\frac{25}{0}$, so we can't evaluate directly. We need to simplify the expression.

Step 2: Factor both numerator and denominator

• The numerator $2x^2 + 4x - 5$ can be factored as: $2x^2 + 4x - 5 = (2x - 1)(x + 5)$

• The denominator $x^2 - 5x + 6$ can be factored as: $x^2 - 5x + 6 = (x - 2)(x - 3)$

Thus, the expression becomes: $\frac{(2x - 1)(x + 5)}{(x - 2)(x - 3)}$

Step 3: Cancel common factors

There are no common factors to cancel, but now we can substitute $x = 3$ directly into the simplified expression:

$\frac{(2(3) - 1)(3 + 5)}{(3 - 2)(3 - 3)} = \frac{(6 - 1)(8)}{(1)(0)} = \frac{5 \cdot 8}{1 \cdot 0}$

This still results in division by zero, implying the limit does not exist. The function approaches infinity or negative infinity as $x$ approaches 3.

Conclusion:

The limit does not exist because the denominator tends to zero while the numerator tends to a non-zero number, indicating that the function has a vertical asymptote at $x = 3$.

Would you like more details or further explanation?

Here are five related questions you might find helpful:

1. What is the definition of a vertical asymptote?
2. How can we determine the behavior of the function as it approaches the asymptote?
3. How do you handle limits involving indeterminate forms like $\frac{0}{0}$?
4. Can we apply L'Hôpital's Rule to this problem, and how would it change the solution?
5. What happens if the numerator also tends to zero when finding limits?

Tip: Always check for factorization when encountering indeterminate forms like $\frac{0}{0}$ in limit problems.