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

• A. $b = 3 \text{ and } L = -1$
• B. $b = -3 \text{ and } L = -1$
• C. $b = 3 \text{ and } L = 1$
• D. $b = -3 \text{ and } L = 1$

Hint:

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:
\[ L = \lim_{x \to -2} \frac{\frac{d}{dx}(3x^2 + 15x + 18)}{\frac{d}{dx}(x^2 + x - 2)} = \lim_{x \to -2} \frac{6x + 15}{2x + 1} = \frac{-12 + 15}{-4 + 1} = -1 \]

Answer 1

The Correct Option is A

Step 1: Check the denominator at $x = -2$.
The bottom is $x^2 + x - 2 = (x + 2)(x - 1)$. At $x = -2$ this is zero. For the whole limit to be a finite number $L$, the top must also be zero at $x = -2$, otherwise the fraction would blow up.

Step 2: Force the numerator to vanish at $x = -2$.
The top is $bx^2 + 15x + 15 + b$. Plug in $x = -2$: $4b - 30 + 15 + b = 5b - 15$. Set this equal to zero: $5b - 15 = 0$, so $b = 3$.

Step 3: Rewrite the numerator with $b = 3$.
The top becomes $3x^2 + 15x + 18 = 3(x^2 + 5x + 6) = 3(x + 2)(x + 3)$.

Step 4: Cancel the common factor.
The fraction is now $\frac{3(x + 2)(x + 3)}{(x + 2)(x - 1)}$. The $(x + 2)$ cancels, leaving $\frac{3(x + 3)}{x - 1}$.

Step 5: Substitute $x = -2$ into the simplified form.
We get $\frac{3(-2 + 3)}{-2 - 1} = \frac{3(1)}{-3} = -1$. So $L = -1$.

Step 6: State both answers together.
Thus $b = 3$ and $L = -1$.
\[ \boxed{b = 3,\ L = -1} \]