Evaluate limx→1−3x^2+5x−2/2x^2−11x+9.

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.

Evaluate limx→1−(3x^2+5x−2)/(2x^2−11x+9).

Learn how to evaluate the limit of the rational function (3x^2 + 5x - 2) / (2x^2 - 11x + 9) as x approaches 1. This detailed solution explores indeterminate forms, vertical asymptotes, and the behavior of the function near x = 1. Ideal for students in Grades 11-12 or introductory calculus courses.

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