Question

Let the slope of the line \( 45x + 5y + 3 = 0 \) be \( 27r_1 + \frac{9r_2}{2} \) for some \( r_1, r_2 \in \mathbb{R} \).Then\[\lim_{x \to 3} \left( \int_3^x \frac{8t^2}{\frac{3r_1 x}{2} - r_2 x^2 - r_1 x^3 - 3x} \, dt \right)\]is equal to _____.

Answer 1

Correct Answer: 12

We are given a limit problem and need to find its value using L'Hopital's Rule. The expression provided is: \[ \lim_{x \to 3} \frac{\frac{3r_2}{2} - r_1 x^3 - 3x}{r_2 x^2 - r_1 x^3 - 3x} \] and it simplifies to \( 12 \) after applying the rule.

Step 1: Solve for \( 6r_1 + r_2 \)

The given equation: \[ 27r_1 + 9r_2 = -9 \quad \Rightarrow \quad 6r_1 + r_2 = -2 \] This equation helps us find the relationship between \( r_1 \) and \( r_2 \).

Step 2: Set up the limit

The limit is given by: \[ \lim_{x \to 3} \frac{ \int_3^2 8t^2 \, dt }{ r_2 x^2 - r_1 x^3 - 3x } \] We apply L'Hopital's Rule to the expression. First, compute the derivatives of the numerator and denominator.

Step 3: Apply L'Hopital's Rule

Using L'Hopital's Rule: \[ \lim_{x \to 3} \frac{ 8x^2 }{ \frac{3r_2}{2} - 2r_2 x - 3r_1 x^2 } \] After differentiating both the numerator and the denominator, we proceed with the computation.

Step 4: Compute the limit

Now, substitute \( x = 3 \) into the expression: \[ \lim_{x \to 3} \frac{ 72 }{ 144 } \] Finally, simplifying the expression gives: \[ \lim_{x \to 3} = 12 \]

Final Answer:

The value of the limit is: \[ \boxed{12} \]