Question:
Suppose $\lim_{x \to -2} \frac{bx^2 + 15x + 15 + b}{x^2 + x - 2} = L$, where $b$ and $L$ are real numbers. Then
Suppose $\lim_{x \to -2} \frac{bx^2 + 15x + 15 + b}{x^2 + x - 2} = L$, where $b$ and $L$ are real numbers. Then
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For any limit of the form $\frac{0}{0}$, once you establish the value of the unknown parameter, you can also use L'Hôpital's Rule to find the limit $L$ quickly:
\[ L = \lim_{x \to -2} \frac{\frac{d}{dx}(3x^2 + 15x + 18)}{\frac{d}{dx}(x^2 + x - 2)} = \lim_{x \to -2} \frac{6x + 15}{2x + 1} = \frac{-12 + 15}{-4 + 1} = -1 \]
\[ L = \lim_{x \to -2} \frac{\frac{d}{dx}(3x^2 + 15x + 18)}{\frac{d}{dx}(x^2 + x - 2)} = \lim_{x \to -2} \frac{6x + 15}{2x + 1} = \frac{-12 + 15}{-4 + 1} = -1 \]
Updated On: Jun 16, 2026
- $b = 3 \text{ and } L = -1$
- $b = -3 \text{ and } L = -1$
- $b = 3 \text{ and } L = 1$
- $b = -3 \text{ and } L = 1$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem presents a limit of a rational function as $x \to -2$.
Since the limit is a finite real number $L$, we need to analyze the behavior of the numerator and denominator near $x = -2$ to find the constants $b$ and $L$.
Step 2: Key Formula or Approach:
If $\lim_{x \to c} \frac{P(x)}{Q(x)} = L$ (where $L$ is finite) and $\lim_{x \to c} Q(x) = 0$, then we must have $\lim_{x \to c} P(x) = 0$ for a valid limit to exist.
We will use factoring of polynomials to evaluate the indeterminate form.
Step 3: Detailed Explanation:
• First, let us check the limit of the denominator as $x \to -2$:
\[ \lim_{x \to -2} (x^2 + x - 2) = (-2)^2 + (-2) - 2 = 4 - 2 - 2 = 0 \]
• Since the denominator goes to 0 and the limit $L$ is a real number, the numerator must also go to 0 as $x \to -2$.
Thus, $x = -2$ must be a root of the numerator $P(x) = bx^2 + 15x + 15 + b$:
\[ P(-2) = 0 \implies b(-2)^2 + 15(-2) + 15 + b = 0 \]
\[ 4b - 30 + 15 + b = 0 \implies 5b - 15 = 0 \implies b = 3 \]
• Now substitute $b = 3$ back into the limit expression:
\[ \lim_{x \to -2} \frac{3x^2 + 15x + 15 + 3}{x^2 + x - 2} = \lim_{x \to -2} \frac{3x^2 + 15x + 18}{x^2 + x - 2} \]
• We can factor both the numerator and the denominator:
Numerator:
\[ 3x^2 + 15x + 18 = 3(x^2 + 5x + 6) = 3(x + 2)(x + 3) \]
Denominator:
\[ x^2 + x - 2 = (x + 2)(x - 1) \]
• Now substitute these factored forms back into the limit:
\[ L = \lim_{x \to -2} \frac{3(x + 2)(x + 3)}{(x + 2)(x - 1)} \]
• Since $x \to -2$, we can cancel the common non-zero term $(x + 2)$:
\[ L = \lim_{x \to -2} \frac{3(x + 3)}{x - 1} \]
• Evaluating the limit directly by substitution:
\[ L = \frac{3(-2 + 3)}{-2 - 1} = \frac{3(1)}{-3} = -1 \]
Step 4: Final Answer:
Therefore, $b = 3$ and $L = -1$.
The problem presents a limit of a rational function as $x \to -2$.
Since the limit is a finite real number $L$, we need to analyze the behavior of the numerator and denominator near $x = -2$ to find the constants $b$ and $L$.
Step 2: Key Formula or Approach:
If $\lim_{x \to c} \frac{P(x)}{Q(x)} = L$ (where $L$ is finite) and $\lim_{x \to c} Q(x) = 0$, then we must have $\lim_{x \to c} P(x) = 0$ for a valid limit to exist.
We will use factoring of polynomials to evaluate the indeterminate form.
Step 3: Detailed Explanation:
• First, let us check the limit of the denominator as $x \to -2$:
\[ \lim_{x \to -2} (x^2 + x - 2) = (-2)^2 + (-2) - 2 = 4 - 2 - 2 = 0 \]
• Since the denominator goes to 0 and the limit $L$ is a real number, the numerator must also go to 0 as $x \to -2$.
Thus, $x = -2$ must be a root of the numerator $P(x) = bx^2 + 15x + 15 + b$:
\[ P(-2) = 0 \implies b(-2)^2 + 15(-2) + 15 + b = 0 \]
\[ 4b - 30 + 15 + b = 0 \implies 5b - 15 = 0 \implies b = 3 \]
• Now substitute $b = 3$ back into the limit expression:
\[ \lim_{x \to -2} \frac{3x^2 + 15x + 15 + 3}{x^2 + x - 2} = \lim_{x \to -2} \frac{3x^2 + 15x + 18}{x^2 + x - 2} \]
• We can factor both the numerator and the denominator:
Numerator:
\[ 3x^2 + 15x + 18 = 3(x^2 + 5x + 6) = 3(x + 2)(x + 3) \]
Denominator:
\[ x^2 + x - 2 = (x + 2)(x - 1) \]
• Now substitute these factored forms back into the limit:
\[ L = \lim_{x \to -2} \frac{3(x + 2)(x + 3)}{(x + 2)(x - 1)} \]
• Since $x \to -2$, we can cancel the common non-zero term $(x + 2)$:
\[ L = \lim_{x \to -2} \frac{3(x + 3)}{x - 1} \]
• Evaluating the limit directly by substitution:
\[ L = \frac{3(-2 + 3)}{-2 - 1} = \frac{3(1)}{-3} = -1 \]
Step 4: Final Answer:
Therefore, $b = 3$ and $L = -1$.
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