For what value of a, if any, does 
limit as x approaches a^-of (x^2+5x+6)/(x-a)=infinity? Be sure to show all work justifying your answer.

We are tasked with determining for which value of \( a \), if any, the limit as \( x \) approaches \( a^{-} \) of \( \frac{x^2 + 5x + 6}{x - a} \) equals infinity. We need to examine this limit carefully and ensure our work is clearly justified.

### Step 1: Express the Limit
The limit we want to compute is:

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

### Step 2: Factor the Numerator
Let's first attempt to factor the numerator \( x^2 + 5x + 6 \). We can factor it as:

\[
x^2 + 5x + 6 = (x + 2)(x + 3)
\]

So the expression becomes:

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

### Step 3: Investigate the Behavior as \( x \to a^{-} \)

For the limit to go to infinity, the denominator \( x - a \) must approach 0 (from the left), while the numerator does not approach 0 at the same rate. This occurs when the value of \( a \) makes the denominator small while the numerator remains non-zero.

### Step 4: Analyze at Critical Points
Let's substitute potential values of \( a \) where the denominator becomes zero, and examine the behavior of the function:

1. **At \( a = -2 \):**

    When \( a = -2 \), the denominator becomes \( x - (-2) = x + 2 \), and the limit becomes:

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

    Simplifying, we cancel out \( x + 2 \) from the numerator and denominator:

    \[
    \lim_{x \to -2^{-}} (x + 3) = -2 + 3 = 1
    \]

    This limit is finite (equal to 1), so \( a = -2 \) does not satisfy the condition for the limit to approach infinity.

2. **At \( a = -3 \):**

    When \( a = -3 \), the denominator becomes \( x - (-3) = x + 3 \), and the limit becomes:

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

    Simplifying by canceling \( x + 3 \) from the numerator and denominator:

    \[
    \lim_{x \to -3^{-}} (x + 2)
    \]

    As \( x \to -3^{-} \), \( x + 2 \to -1 \), which is finite (equal to -1). Therefore, \( a = -3 \) also does not result in an infinite limit.

### Step 5: Conclusion
Neither \( a = -2 \) nor \( a = -3 \) causes the limit to go to infinity. Hence, there is **no value of \( a \)** such that the limit as \( x \to a^{-} \) of \( \frac{x^2 + 5x + 6}{x - a} \) equals infinity.

### Summary:
- We factored the expression and tested critical points \( a = -2 \) and \( a = -3 \), but neither value results in the limit tending to infinity.
- Therefore, there is no value of \( a \) that satisfies the condition for the limit to approach infinity.

Would you like more details on any step? Here are 5 related questions to deepen your understanding:

1. What is the significance of factoring polynomials in limit problems?
2. How does the cancellation of terms affect the limit outcome?
3. What conditions typically cause a limit to approach infinity?
4. How do limits from the left and right differ in behavior?
5. What happens if a discontinuity occurs in the denominator of a limit expression?

**Tip:** When dealing with limits approaching infinity, always check for vertical asymptotes where the denominator tends to zero without canceling with the numerator.

For what value of a, if any, does limit as x approaches a^- of (x^2 + 5x + 6) / (x - a) = infinity? Be sure to show all work justifying your answer.

Explore how to determine the value of a where the limit of (x^2 + 5x + 6) / (x - a) as x approaches a^- results in infinity. This advanced problem involves concepts such as limits, factoring, and asymptotic behavior. Detailed step-by-step solution provided. Suitable for Grades 11-12.

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