Find the indicated limit.

lim *x*→2 

*x*2 + *x* − 6*/x*2 − 5*x* + 6

To find the limit of the expression \(\lim_{x \to 2} \frac{x^2 + x - 6}{x^2 - 5x + 6}\), follow these steps:

### 1. Substitute \(x = 2\) into the Expression
First, let's try to substitute \(x = 2\) directly into the expression:

\[
\frac{2^2 + 2 - 6}{2^2 - 5(2) + 6} = \frac{4 + 2 - 6}{4 - 10 + 6}
\]

Simplifying both the numerator and the denominator:

\[
\text{Numerator: } 4 + 2 - 6 = 0
\]
\[
\text{Denominator: } 4 - 10 + 6 = 0
\]

This results in the indeterminate form \(\frac{0}{0}\), so direct substitution does not work. We need to simplify the expression first.

### 2. Simplify the Expression by Factorization
From the previous solution, we know that:

\[
\frac{x^2 + x - 6}{x^2 - 5x + 6} = \frac{(x + 3)(x - 2)}{(x - 3)(x - 2)}
\]

After canceling the common factor \((x - 2)\) (as long as \(x \neq 2\)):

\[
\frac{x + 3}{x - 3}
\]

### 3. Find the Limit of the Simplified Expression
Now, find the limit as \(x\) approaches 2 for the simplified expression:

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

Substitute \(x = 2\) into the simplified expression:

\[
\frac{2 + 3}{2 - 3} = \frac{5}{-1} = -5
\]

### Conclusion
Thus, the limit is:

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

Would you like further details or have any questions?

Here are 5 related questions:
1. How do you handle limits that result in indeterminate forms like \(\frac{0}{0}\)?
2. What are the different types of indeterminate forms and how can they be resolved?
3. Can L'Hôpital's Rule be applied to this limit problem? Why or why not?
4. How do you determine if a function is continuous at a point?
5. What are the implications of canceling out terms when evaluating limits?

**Tip:** When you encounter an indeterminate form like \(\frac{0}{0}\), factorization or other algebraic techniques can often simplify the expression enough to find the limit.

Learn how to find the limit of the expression (x^2 + x - 6) / (x^2 - 5x + 6) as x approaches 2 with a detailed step-by-step solution. This problem covers key concepts in limits, including handling indeterminate forms and using algebraic techniques like factorization. Suitable for students in Grades 11-12.

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