mathop {lim }limits_{n to infty } {left[ {fra | vedclass

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(C) Let $L = \mathop {\lim }\limits_{n 	o \infty } {\left[ {\frac{{\sin n}}{{{n^2}}} + \log \left( {\frac{{en + 1}}{{n + e}}} ight)} ight]^n}$.As

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$e^{\frac{1}{e} - e}$

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(C) Let $L = \mathop {\lim }\limits_{n 	o \infty } {\left[ {\frac{{\sin n}}{{{n^2}}} + \log \left( {\frac{{en + 1}}{{n + e}}} ight)} ight]^n}$.As $n 	o \infty$,$\frac{\sin n}{n^2} 	o 0$ and $\log \left( \frac{en+1}{n+e} ight) 	o \log(e) = 1$. Thus,it is a $1^\infty$ form.Using the formula $\lim_{n 	o \infty} [f(n)]^{g(n)} = e^{\lim_{n 	o \infty} g(n)(f(n)-1)}$,we have:$L = e^{\mathop {\lim }\limits_{n 	o \infty } n \left[ {\frac{{\sin n}}{{{n^2}}} + \log \left( {\frac{{en + 1}}{{n + e}}} ight) - 1} ight]}$$= e^{\mathop {\lim }\limits_{n 	o \infty } \left[ {\frac{{\sin n}}{n} + n \log \left( {\frac{{en + 1}}{{n + e}}} ight) - n} ight]}$$= e^{\mathop {\lim }\limits_{n 	o \infty } \left[ {\frac{{\sin n}}{n} + n \log \left( \frac{en+1}{n+e} ight) - n \log e} ight]}$$= e^{\mathop {\lim }\limits_{n 	o \infty } \left[ {\frac{{\sin n}}{n} + n \log \left( \frac{en+1}{e(n+e)} ight)} ight]}$$= e^{\mathop {\lim }\limits_{n 	o \infty } \left[ {\frac{{\sin n}}{n} + n \log \left( \frac{en+1}{en+e^2} ight)} ight]}$$= e^{\mathop {\lim }\limits_{n 	o \infty } \left[ {\frac{{\sin n}}{n} + n \log \left( 1 + \frac{1-e^2}{en+e^2} ight)} ight]}$Since $\frac{\sin n}{n} 	o 0$ and $n \log(1+x) \approx nx$ as $x 	o 0$:$= e^{0 + \lim_{n 	o \infty} n \cdot \frac{1-e^2}{en+e^2}} = e^{\frac{1-e^2}{e}} = e^{\frac{1}{e} - e}$.

$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{\sin \left( n \right)}}{{{n^2}}} + \log \left( {\frac{{en + 1}}{{n + e}}} \right)} \right]^n}$ is equal to

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