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: Understanding the Concept
This is a limit problem that evaluates to the indeterminate form \(\frac{0}{0}\) upon direct substitution. We can solve it by factoring the algebraic expressions and using the standard limit \(\lim_{u \to 0} \frac{\sin u}{u} = 1\).
Step 2: Key Formula or Approach
1. Factor the numerator: Use the difference of cubes formula, \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\).
2. Factor the denominator: Recognize it as a perfect square trinomial.
3. Rearrange the expression to isolate a \(\frac{\sin(x-2)}{x-2}\) term.
4. Evaluate the limit of the remaining parts.
Step 3: Detailed Explanation
1. Factor the algebraic parts of the expression.
The limit is \(L = \lim_{x \to 2} \frac{(x^3 - 8)\sin(x-2)}{x^2 - 4x + 4}\).
- Numerator: \(x^3 - 8 = x^3 - 2^3 = (x-2)(x^2 + 2x + 4)\).
- Denominator: \(x^2 - 4x + 4 = (x-2)^2\).
2. Substitute the factored forms back into the limit.
\[ L = \lim_{x \to 2} \frac{(x-2)(x^2 + 2x + 4)\sin(x-2)}{(x-2)^2} \] 3. Simplify and rearrange the expression.
Cancel one \((x-2)\) term from the numerator and denominator.
\[ L = \lim_{x \to 2} \frac{(x^2 + 2x + 4)\sin(x-2)}{x-2} \] Group the terms to apply the standard limit:
\[ L = \lim_{x \to 2} \left[ (x^2 + 2x + 4) \cdot \frac{\sin(x-2)}{x-2} \right] \] 4. Evaluate the limits of the individual parts.
Since the limits of both parts exist, we can evaluate them separately.
- First part: \[ \lim_{x \to 2} (x^2 + 2x + 4) = (2)^2 + 2(2) + 4 = 4 + 4 + 4 = 12 \] - Second part: Let \(u = x-2\). As \(x \to 2\), \(u \to 0\). \[ \lim_{x \to 2} \frac{\sin(x-2)}{x-2} = \lim_{u \to 0} \frac{\sin u}{u} = 1 \] 5. Multiply the results.
\[ L = (12) \times (1) = 12 \] Step 4: Final Answer
The value of the limit is 12.