Question:medium
The value of \(\lim_{x \to 0} \frac{6\sin(x) - 2\sin(3x)}{3x^3}\) is equal to
The value of \(\lim_{x \to 0} \frac{6\sin(x) - 2\sin(3x)}{3x^3}\) is equal to
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Taylor expansions are often faster than applying L'Hopital's rule three times for cubic denominators. Always expand up to the same power as the highest power in the denominator.
Updated On: Jun 25, 2026
- \(-\frac{8}{3}\)
- \(\frac{8}{3}\)
- \(-\frac{8}{9}\)
- \(\frac{8}{9}\)
- 0
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The Correct Option is B
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