Question

\(\displaystyle\lim_{x\rightarrow-2}\frac{3x^2+ax-2}{x^2-x-6}\) is a finite number, then the value of a is equal to

• A. 2
• B. 3
• C. 4
• D. 5
• E. 6

Answer 1

The Correct Option is D

We are given the limit: \[ \lim_{x \to 2} \frac{3x^2 + ax - 2}{x^2 - x - 6} \] First, factor the denominator: \[ x^2 - x - 6 = (x - 3)(x + 2) \] Now, substitute \( x = 2 \) into the denominator: \[ (x - 3)(x + 2) \bigg|_{x = 2} = (2 - 3)(2 + 2) = (-1)(4) = -4 \] For the limit to be finite, the numerator must also approach zero at \( x = 2 \), otherwise, the limit would be infinite. Therefore, substitute \( x = 2 \) into the numerator: \[ 3x^2 + ax - 2 \bigg|_{x = 2} = 3(2^2) + a(2) - 2 = 12 + 2a - 2 = 10 + 2a \] For the numerator to approach zero at \( x = 2 \), we set: \[ 10 + 2a = 0 \] Solving for \( a \): \[ 2a = -10 \quad \Rightarrow \quad a = -5 \]

The correct option is (D) : \(5\)

Answer 2

• We are given the limit: \[\displaystyle\lim_{x\rightarrow-2}\frac{3x^2+ax-2}{x^2-x-6}\] and we want it to be a finite number.
• First, factor the denominator: \[x^2 - x - 6 = (x-3)(x+2)\]
• Since the limit exists as \(x\) approaches -2, the numerator must also be zero at \(x = -2\) to avoid a singularity of the form \(\frac{k}{0}\) where k is a non-zero number. Otherwise, the limit would be infinite.
• Therefore, we must have: \[3(-2)^2 + a(-2) - 2 = 0\]
• Simplify the equation: \[3(4) - 2a - 2 = 0\] \[12 - 2a - 2 = 0\] \[10 - 2a = 0\]
• Solve for \(a\): \[2a = 10\] \[a = 5\]
• Now, with \(a = 5\), the numerator becomes: \[3x^2 + 5x - 2\]
• Factor the numerator: \[3x^2 + 5x - 2 = (3x-1)(x+2)\]
• So, the limit becomes: \[\displaystyle\lim_{x\rightarrow-2}\frac{(3x-1)(x+2)}{(x-3)(x+2)} = \displaystyle\lim_{x\rightarrow-2}\frac{3x-1}{x-3}\]
• Evaluate the limit by plugging in \(x = -2\): \[\frac{3(-2) - 1}{-2 - 3} = \frac{-6 - 1}{-5} = \frac{-7}{-5} = \frac{7}{5}\]
• Since the limit is \(\frac{7}{5}\), which is a finite number, the value of \(a\) is 5.