If mathop {lim }limits_{x to 0} left( {frac{{ | vedclass

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(B) We use the Taylor series expansion for $\sin x$: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = x - \frac{x^3}{6} + \frac{x^5}{120} - \do

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$5$

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(B) We use the Taylor series expansion for $\sin x$: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$Substituting this into the expression:$\mathop {\lim }\limits_{x 	o 0} \frac{3(x - \frac{x^3}{6} + \frac{x^5}{120} - \dots) - 3x + \frac{x^3}{2}}{2x^n}$$= \mathop {\lim }\limits_{x 	o 0} \frac{3x - \frac{3x^3}{6} + \frac{3x^5}{120} - 3x + \frac{x^3}{2}}{2x^n}$$= \mathop {\lim }\limits_{x 	o 0} \frac{-\frac{x^3}{2} + \frac{x^5}{40} + \frac{x^3}{2}}{2x^n}$$= \mathop {\lim }\limits_{x 	o 0} \frac{\frac{x^5}{40}}{2x^n} = \mathop {\lim }\limits_{x 	o 0} \frac{x^5}{80x^n}$For the limit to be a finite non-zero number,the power of $x$ in the denominator must match the power of $x$ in the numerator.Thus,$n = 5$.

If $\mathop {\lim }\limits_{x \to 0} \left( {\frac{{3\sin x - 3x + \frac{{{x^3}}}{2}}}{{2{x^n}}}} \right)$ is a finite number,then the greatest value of $n \in N$ is -

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