Question:medium
$\lim_{x\rightarrow2}\frac{(x^{3}-8)\sin(x-2)}{x^{2}-4x+4}$ is equal to ________.
$\lim_{x\rightarrow2}\frac{(x^{3}-8)\sin(x-2)}{x^{2}-4x+4}$ is equal to ________.
Show Hint
$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.
Updated On: Jun 24, 2026
- 4
- 8
- 12
- -8
- -12
Show Solution
The Correct Option is C
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