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

Step 1: Direct substitution
Substitute $x = 2$: \[ \text{Numerator} = 2^3 + 3(2)^2 - 9(2) - 2 = 8 + 12 - 18 - 2 = 0 \] \[ \text{Denominator} = 2^3 - (2)^2 - 4(2) + 4 = 8 - 4 - 8 + 4 = 0 \] Thus, the limit is of the form $\frac{0}{0}$ (indeterminate).
Step 2: Apply L'Hôpital's Rule
Differentiate numerator and denominator: \[ f'(x) = 3x^2 + 6x - 9, g'(x) = 3x^2 - 2x - 4 \]
Step 3: Evaluate the new limit
\[ \lim_{x \to 2} \frac{3x^2 + 6x - 9}{3x^2 - 2x - 4} \] Substitute $x = 2$: \[ = \frac{3(2)^2 + 6(2) - 9}{3(2)^2 - 2(2) - 4} = \frac{12 + 12 - 9}{12 - 4 - 4} = \frac{15}{4} \] Final Answer:
\[ \boxed{\frac{15}{4}} \]