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{:[=b ln sqrt(x+1)-ln sqrt1],[=p ln sqrtoo]:} Evaluate the following Limit using Maclaur in Series: lim_(x rarr0)(x+(x^(3))/(3)-sin x)/(2x^(3))

=bln⁡x+1−ln⁡1=pln⁡∞ \begin{array}{l} =b \ln \sqrt{x+1}-\ln \sqrt{1} \\ =p \ln \sqrt{\infty} \end{array} \newlineEvaluate the following Limit using Maclaur in Series:\newlinelim⁡x→0x+x33−sin⁡x2x3 \lim _{x \rightarrow 0} \frac{x+\frac{x^{3}}{3}-\sin x}{2 x^{3}}

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Q. =bln⁡x+1−ln⁡1=pln⁡∞ \begin{array}{l} =b \ln \sqrt{x+1}-\ln \sqrt{1} \\ =p \ln \sqrt{\infty} \end{array} \newlineEvaluate the following Limit using Maclaur in Series:\newlinelim⁡x→0x+x33−sin⁡x2x3 \lim _{x \rightarrow 0} \frac{x+\frac{x^{3}}{3}-\sin x}{2 x^{3}}
  1. Identify function: Identify the function to analyze:\newlineWe have the function (x+x33−sin⁡x)/(2x3)(x + \frac{x^3}{3} - \sin x) / (2x^3).
  2. Use Taylor series: Use Taylor series expansion for sin⁡x\sin x around 00:\newlinesin⁡x≈x−x36+O(x5)\sin x \approx x - \frac{x^3}{6} + O(x^5) as x→0x \to 0.
  3. Substitute series into function: Substitute the Taylor series into the function:\newlineReplace sin⁡x\sin x in the function:\newlinex+x33−(x−x36)2x3=x+x33−x+x362x3 \frac{x + \frac{x^3}{3} - (x - \frac{x^3}{6})}{2x^3} = \frac{x + \frac{x^3}{3} - x + \frac{x^3}{6}}{2x^3}
  4. Simplify numerator: Simplify the numerator:\newlineCombine like terms in the numerator:\newlinex33+x362x3=2x36+x362x3=3x362x3=x322x3 \frac{\frac{x^3}{3} + \frac{x^3}{6}}{2x^3} = \frac{\frac{2x^3}{6} + \frac{x^3}{6}}{2x^3} = \frac{\frac{3x^3}{6}}{2x^3} = \frac{\frac{x^3}{2}}{2x^3}
  5. Further simplify expression: Further simplify the expression:\newlinex322x3=14 \frac{\frac{x^3}{2}}{2x^3} = \frac{1}{4}

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