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.