Question

$\lim_{x\rightarrow2}\frac{(x^{3}-8)\sin(x-2)}{x^{2}-4x+4}$ is equal to ________.

• A. 4
• B. 8
• C. 12
• D. -8
• E. -12

Hint:

$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

Answer 1

The Correct Option is C

Step 1: Concept
Simplify the expression using algebraic identities and standard limits.

Step 2: Meaning
Factorize: $x^3-8 = (x-2)(x^2+2x+4)$ and $x^2-4x+4 = (x-2)^2$.

Step 3: Analysis
The limit becomes: $\lim_{x\rightarrow2} \frac{(x-2)(x^2+2x+4)\sin(x-2)}{(x-2)^2} = \lim_{x\rightarrow2} (x^2+2x+4) \cdot \frac{\sin(x-2)}{x-2}$.

Step 4: Conclusion
As $x \to 2$, $\frac{\sin(x-2)}{x-2} \to 1$ and $(x^2+2x+4) \to (4+4+4) = 12$. Final Answer: (C)