Question

If \( \lim_{x \to 4} \frac{2x^2 + (3+2a)x + 3a{x^3 - 2x^2 - 23x + 60} = \frac{11}{9} \), then find \( \lim_{x \to a} \frac{x^2 + 9x + 20}{x^2 - x - 20} \).}

• A. \( -9 \)
• B. \( -4 \)
• C. \( -\frac{1}{4} \)
• D. \( -\frac{1}{9} \) \bigskip

Hint:

When you encounter a limit with indeterminate form \( \frac{0}{0} \), apply L'Hopital's Rule by differentiating both the numerator and denominator. Then evaluate the limit.

Answer 1

The Correct Option is D

We are given the following limit: \[ \lim_{x \to 4} \frac{2x^2 + (3 + 2a)x + 3a}{x^3 - 2x^2 - 23x + 60} = \frac{11}{9} \] Step 1: Simplifying the denominator at \( x = 4 \) First, factor the denominator \( x^3 - 2x^2 - 23x + 60 \). Start by trying \( x = 4 \) as a root: \[ 4^3 - 2(4)^2 - 23(4) + 60 = 64 - 32 - 92 + 60 = 0 \] Thus, \( x = 4 \) is a root. We can divide the cubic polynomial by \( (x - 4) \) using synthetic division: \[ \frac{x^3 - 2x^2 - 23x + 60}{x - 4} = x^2 + 2x - 15 \] Thus, the denominator can be factored as: \[ x^3 - 2x^2 - 23x + 60 = (x - 4)(x^2 + 2x - 15) \] Step 2: Substituting in the given limit Now, we can substitute the denominator into the given expression: \[ \lim_{x \to 4} \frac{2x^2 + (3 + 2a)x + 3a}{(x - 4)(x^2 + 2x - 15)} = \frac{11}{9} \] Since \( x = 4 \) is a root of the denominator, the numerator must also be zero when \( x = 4 \). Therefore, substitute \( x = 4 \) into the numerator: \[ 2(4)^2 + (3 + 2a)(4) + 3a = 0 \] \[ 32 + 4(3 + 2a) + 3a = 0 \] \[ 32 + 12 + 8a + 3a = 0 \implies 44 + 11a = 0 \] \[ a = -\frac{44}{11} = -4 \] Step 3: Substituting \( a = -4 \) in the second limit Now that we know \( a = -4 \), we need to find: \[ \lim_{x \to a} \frac{x^2 + 9x + 20}{x^2 - x - 20} = \lim_{x \to -4} \frac{x^2 + 9x + 20}{x^2 - x - 20} \] Substitute \( x = -4 \) into the numerator and denominator: For the numerator: \[ (-4)^2 + 9(-4) + 20 = 16 - 36 + 20 = 0 \] For the denominator: \[ (-4)^2 - (-4) - 20 = 16 + 4 - 20 = 0 \] Thus, both the numerator and denominator are zero, so we apply L'Hopital's Rule. First, take the derivatives of the numerator and denominator: Numerator's derivative: \[ \frac{d}{dx} (x^2 + 9x + 20) = 2x + 9 \] Denominator's derivative: \[ \frac{d}{dx} (x^2 - x - 20) = 2x - 1 \] Now, substitute \( x = -4 \) into these derivatives: \[ \text{Numerator: } 2(-4) + 9 = -8 + 9 = 1 \] \[ \text{Denominator: } 2(-4) - 1 = -8 - 1 = -9 \] Thus, the limit is: \[ \lim_{x \to -4} \frac{2x + 9}{2x - 1} = \frac{1}{-9} = -\frac{1}{9} \] Thus, the value of the limit is \( -\frac{1}{9} \). \bigskip