Question

The value of $\lim_{x \to 2} \frac{x^3 + 3x^2 - 9x - 2}{x^3 - x^2 - 4x + 4}$ is ________

• A. 3
• B. $\frac{15}{4}$
• C. $\frac{15}{2}$
• D. $\frac{15}{13}$

Hint:

L'Hôpital's Rule is a powerful tool, but always remember to verify that the limit is in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying it. Applying it to a determinate form will yield an incorrect result.

Answer 1

The Correct Option is B

To find the value of the limit \(\lim_{x \to 2} \frac{x^3 + 3x^2 - 9x - 2}{x^3 - x^2 - 4x + 4}\), we start by evaluating the expression directly at \(x = 2\). However, direct substitution might lead to a \(\frac{0}{0}\) indeterminate form, so we need to factor the polynomials and simplify.

Step-by-step Solution:

1. First, substitute \(x = 2\) into both the numerator \(x^3 + 3x^2 - 9x - 2\) and the denominator \(x^3 - x^2 - 4x + 4\):

• Substituting \(x = 2\) into the numerator: \((2)^3 + 3(2)^2 - 9(2) - 2 = 8 + 12 - 18 - 2 = 0\).
• Substituting \(x = 2\) into the denominator: \((2)^3 - (2)^2 - 4(2) + 4 = 8 - 4 - 8 + 4 = 0\).

Both the numerator and the denominator evaluate to zero, confirming the indeterminate form \(\frac{0}{0}\).

1. Next, factor both the numerator and the denominator to cancel out common factors:

• The numerator \(x^3 + 3x^2 - 9x - 2\) can be factored as \((x - 2)(x^2 + 5x + 1)\).
• The denominator \(x^3 - x^2 - 4x + 4\) can be factored as \((x - 2)(x^2 + x - 2)\).

1. Cancel the common factor \((x - 2)\) from both the numerator and the denominator:

The expression simplifies to \(\frac{x^2 + 5x + 1}{x^2 + x - 2}\).

1. Now, evaluate the limit as \(x\) approaches 2:

• Substituting \(x = 2\) in the simplified expression: \(\frac{(2)^2 + 5(2) + 1}{(2)^2 + 2 - 2} = \frac{4 + 10 + 1}{4 + 2 - 2} = \frac{15}{4}\).

Thus, the value of the limit is \(\frac{15}{4}\).

Conclusion:

The correct answer is \(\frac{15}{4}\).