Let a_n denote the term independent of x in t | vedclass

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(A) The general term in the expansion of $\left[x+\frac{\sin(1/n)}{x^2}\right]^{3n}$ is given by $T_{r+1} = {}^{3n}C_r (x)^{3n-r} \left(\frac{\sin(1/n

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(A) The general term in the expansion of $\left[x+\frac{\sin(1/n)}{x^2}ight]^{3n}$ is given by $T_{r+1} = {}^{3n}C_r (x)^{3n-r} \left(\frac{\sin(1/n)}{x^2}ight)^r = {}^{3n}C_r (x)^{3n-3r} (\sin(1/n))^r$. For the term to be independent of $x$, the exponent of $x$ must be zero, so $3n - 3r = 0$, which implies $r = n$. Thus, $a_n = {}^{3n}C_n (\sin(1/n))^n$. We need to evaluate $\lim_{n 	o \infty} \frac{a_n \cdot n!}{^{3n}P_n}$. Since ${}^{3n}P_n = \frac{(3n)!}{(3n-n)!} = \frac{(3n)!}{(2n)!}$, we have $\frac{a_n \cdot n!}{^{3n}P_n} = \frac{{}^{3n}C_n (\sin(1/n))^n \cdot n! \cdot (2n)!}{(3n)!} = \frac{(3n)!}{n!(2n)!} \cdot (\sin(1/n))^n \cdot \frac{n!(2n)!}{(3n)!} = (\sin(1/n))^n$. As $n 	o \infty$, $\sin(1/n) \approx 1/n$. Thus, $\lim_{n 	o \infty} (\sin(1/n))^n = \lim_{n 	o \infty} (1/n)^n = 0$.

Let $a_n$ denote the term independent of $x$ in the expansion of $\left[x+\frac{\sin(1/n)}{x^2}\right]^{3n}$. Then $\lim_{n \to \infty} \frac{a_n \cdot n!}{^{3n}P_n}$ equals

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