Find the number a such that the limit exists.
 
lim 
x→−2 
4x2 + ax + a + 12
x2 + x − 2
 
a = 
 and Find the value of the limit

To find the value of \( a \) such that the limit

\[
\lim_{x \to -2} \frac{4x^2 + ax + a + 12}{x^2 + x - 2}
\]

exists, we will need to ensure that the denominator does not approach zero unless the numerator also does, which would allow us to potentially simplify the expression.

### Step 1: Factor the denominator

We first factor the denominator \( x^2 + x - 2 \). Solving \( x^2 + x - 2 = 0 \), we factor it as follows:

\[
x^2 + x - 2 = (x - 1)(x + 2)
\]

### Step 2: Analyze the behavior of the denominator at \( x = -2 \)

At \( x = -2 \), the denominator becomes zero:

\[
x^2 + x - 2 = (-2 - 1)(-2 + 2) = (-3)(0) = 0
\]

For the limit to exist, the numerator must also go to zero when \( x = -2 \), so we set up the condition for the numerator.

### Step 3: Set up the numerator

The numerator is \( 4x^2 + ax + a + 12 \). To ensure that the numerator is zero at \( x = -2 \), we substitute \( x = -2 \) into the numerator:

\[
4(-2)^2 + a(-2) + a + 12 = 16 - 2a + a + 12 = 28 - a
\]

For the limit to exist, we require \( 28 - a = 0 \), which gives:

\[
a = 28
\]

### Step 4: Simplify the expression

Now that we have \( a = 28 \), substitute this value back into the numerator:

\[
4x^2 + 28x + 28 + 12 = 4x^2 + 28x + 40
\]

We now rewrite the limit:

\[
\lim_{x \to -2} \frac{4x^2 + 28x + 40}{x^2 + x - 2} = \lim_{x \to -2} \frac{4x^2 + 28x + 40}{(x - 1)(x + 2)}
\]

### Step 5: Factor the numerator

To simplify further, we attempt to factor the numerator \( 4x^2 + 28x + 40 \). Factoring out the greatest common factor:

\[
4x^2 + 28x + 40 = 4(x^2 + 7x + 10)
\]

Next, factor \( x^2 + 7x + 10 \):

\[
x^2 + 7x + 10 = (x + 5)(x + 2)
\]

Thus, the numerator becomes:

\[
4(x + 5)(x + 2)
\]

### Step 6: Cancel common factors

The limit expression is now:

\[
\lim_{x \to -2} \frac{4(x + 5)(x + 2)}{(x - 1)(x + 2)}
\]

We can cancel the common factor \( (x + 2) \) from the numerator and denominator:

\[
\lim_{x \to -2} \frac{4(x + 5)}{x - 1}
\]

### Step 7: Evaluate the limit

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

\[
\frac{4(-2 + 5)}{-2 - 1} = \frac{4(3)}{-3} = \frac{12}{-3} = -4
\]

Thus, the value of the limit is \( -4 \).

### Final Answer

- The value of \( a \) is \( 28 \).
- The value of the limit is \( -4 \).

---

Would you like more details or have any questions?

Here are 5 related questions you can explore:
1. How do you factor quadratic expressions in general?
2. What happens if you cannot cancel common factors in a limit problem?
3. How do you evaluate limits involving indeterminate forms like \( \frac{0}{0} \)?
4. Can you solve limits using L'Hôpital's Rule?
5. What is the geometric interpretation of limits approaching a certain point?

**Tip**: When factoring polynomials, always check for common factors first—it simplifies the work considerably.

Find the number a such that the limit exists.

lim
x→−2
(4x^2 + ax + a + 12) / (x^2 + x − 2)

a = ? and find the value of the limit.

Learn how to find the value of 'a' such that the limit lim x→−2 (4x^2 + ax + a + 12) / (x^2 + x − 2) exists. This problem covers limits and quadratic factorization, suitable for students in Grades 10-12. Explore a step-by-step solution to ensure the numerator and denominator both approach zero and simplify the limit expression.

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