To evaluate the limit

$\lim_{{x \to 1}} \frac{3x^2 + 5x - 2}{2x^2 - 11x + 9},$

we first need to check if direct substitution gives a determinate form or an indeterminate form like $\frac{0}{0}$.

Step 1: Substitute $x = 1$

Let's substitute $x = 1$ into the numerator and the denominator:

Numerator:

$3(1)^2 + 5(1) - 2 = 3 + 5 - 2 = 6.$

Denominator:

$2(1)^2 - 11(1) + 9 = 2 - 11 + 9 = 0.$

Substitution shows that the denominator is $0$, but the numerator is not zero. Therefore, we have a situation of $\frac{6}{0}$, which indicates that the function has a vertical asymptote at $x = 1$.

Conclusion

Since the denominator approaches $0$ while the numerator approaches $6$, the limit is of the form $\frac{\text{non-zero}}{0}$, leading to the conclusion that the limit does not exist in a finite sense. The function has a vertical asymptote at $x = 1$.

Final Answer

$\lim_{{x \to 1}} \frac{3x^2 + 5x - 2}{2x^2 - 11x + 9} = \infty \text{ or } -\infty.$

Would you like me to investigate the one-sided limits to determine whether the limit goes to $\infty$ or $-\infty$ from each side?

Related Questions

1. What is the definition of a vertical asymptote in calculus?
2. How do you evaluate one-sided limits?
3. How can you determine if a function is continuous at a given point?
4. What are indeterminate forms in calculus?
5. How can factoring help in finding limits?

Tip: When dealing with limits that result in an indeterminate form, consider factoring, rationalizing, or applying L'Hopital's Rule to simplify the expression.