It seems like you're trying to express a mathematical limit. I'll interpret your input as:

\[
\lim_{{x \to 2}} \frac{2x^2 - 11x + 14}{x^2 - 5x + 6}
\]

Let's solve this limit.

### Step 1: Factor the numerator and denominator

First, factor both the numerator and the denominator:

- Numerator: \(2x^2 - 11x + 14\)
- Denominator: \(x^2 - 5x + 6\)

The denominator factors easily:
\[
x^2 - 5x + 6 = (x - 2)(x - 3)
\]

Now let's factor the numerator:
We need to factor \(2x^2 - 11x + 14\). Using factoring techniques or the quadratic formula:

\[
2x^2 - 11x + 14 = (2x - 7)(x - 2)
\]

So the expression becomes:

\[
\lim_{{x \to 2}} \frac{(2x - 7)(x - 2)}{(x - 2)(x - 3)}
\]

### Step 2: Cancel common factors

Both the numerator and denominator have a common factor of \((x - 2)\), so we can cancel them out:

\[
\lim_{{x \to 2}} \frac{2x - 7}{x - 3}
\]

### Step 3: Substitute \(x = 2\)

Now substitute \(x = 2\) into the simplified expression:

\[
\frac{2(2) - 7}{2 - 3} = \frac{4 - 7}{2 - 3} = \frac{-3}{-1} = 3
\]

### Final Answer:
\[
\lim_{{x \to 2}} \frac{2x^2 - 11x + 14}{x^2 - 5x + 6} = 3
\]

Would you like a more detailed explanation or have any further questions?

### Related Questions:
1. What are the methods to factor quadratic equations?
2. How do you solve limits involving indeterminate forms?
3. What are other techniques to evaluate limits if factoring doesn't work?
4. How does the quadratic formula help in factoring difficult quadratics?
5. How can L'Hôpital's Rule be applied to limits involving indeterminate forms?

### Tip:
When evaluating limits, always check if factoring or simplifying the expression can help resolve indeterminate forms like \(\frac{0}{0}\).

lim x -> 2 of (2x^2 - 11x + 14) / (x^2 - 5x + 6)

This page provides a step-by-step solution for solving the limit lim x -> 2 of (2x^2 - 11x + 14) / (x^2 - 5x + 6) by factoring both the numerator and denominator. Key concepts include limits, factoring quadratics, and limit laws. Ideal for Pre-Calculus students in Grades 11-12.

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