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Multiple Choice
Use series to evaluate the limit:
A
B
C
D
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1
Step 1: Recall the Taylor series expansion for sin(x) around x = 0: sin(x) = x - \(\frac{x^3}{6}\) + \(\frac{x^5}{120}\) + \(\dots\). Substitute 3x into this expansion to get sin(3x) = 3x - \(\frac{(3x)^3}{6}\) + \(\frac{(3x)^5}{120}\) + \(\dots\).
Step 2: Expand sin(3x) using the Taylor series: sin(3x) = 3x - \(\frac{27x^3}{6}\) + \(\frac{243x^5}{120}\) + \(\dots\). Simplify the coefficients: sin(3x) = 3x - \(\frac{9x^3}{2}\) + \(\frac{81x^5}{40}\) + \(\dots\).
Step 3: Substitute sin(3x) into the numerator of the given limit: \(\sin\)(3x) - 3x + \(\frac{9}{2}\)x^3 = (3x - \(\frac{9x^3}{2}\) + \(\frac{81x^5}{40}\) + \(\dots\)) - 3x + \(\frac{9}{2}\)x^3.
Step 4: Simplify the numerator by combining like terms: \(\sin\)(3x) - 3x + \(\frac{9}{2}\)x^3 = -\(\frac{9x^3}{2}\) + \(\frac{9}{2}\)x^3 + \(\frac{81x^5}{40}\) + \(\dots\) = \(\frac{81x^5}{40}\) + \(\dots\).
Step 5: Divide the simplified numerator by x^5: \(\frac{\frac{81x^5}{40}\) + \(\dots\)}{x^5} = \(\frac{81}{40}\) + \(\dots\). As x approaches 0, higher-order terms vanish, leaving \(\frac{81}{40}\). Simplify further to \(\frac{27}{20}\).