The value of mathop {Limit}limits_{x to infty | vedclass

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(B) Let $L = \mathop {Limit}\limits_{x 	o \infty } \frac{{2^{\frac{x^n}{e^x}}} - {3^{\frac{x^n}{e^x}}}}{x^n}$.Since $\mathop {Limit}\limits_{x 	o \i

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(B) Let $L = \mathop {Limit}\limits_{x 	o \infty } \frac{{2^{\frac{x^n}{e^x}}} - {3^{\frac{x^n}{e^x}}}}{x^n}$.Since $\mathop {Limit}\limits_{x 	o \infty } \frac{x^n}{e^x} = 0$ for any $n \in N$,we can substitute $u = \frac{x^n}{e^x}$,where $u 	o 0$ as $x 	o \infty$.Then $L = \mathop {Limit}\limits_{u 	o 0} \frac{2^u - 3^u}{x^n}$.Since $x^n = u \cdot e^x$,we have $L = \mathop {Limit}\limits_{x 	o \infty } \frac{2^u - 3^u}{u \cdot e^x} = \mathop {Limit}\limits_{u 	o 0} \left( \frac{2^u - 3^u}{u} ight) \cdot \mathop {Limit}\limits_{x 	o \infty } \frac{1}{e^x}$.Using the standard limit $\mathop {Limit}\limits_{u 	o 0} \frac{a^u - 1}{u} = \ln a$,we get $\mathop {Limit}\limits_{u 	o 0} \frac{2^u - 3^u}{u} = \ln 2 - \ln 3 = \ln \left( \frac{2}{3} ight)$.As $x 	o \infty$,$e^x 	o \infty$,so $\frac{1}{e^x} 	o 0$.Therefore,$L = \ln \left( \frac{2}{3} ight) \cdot 0 = 0$.

The value of $\mathop {Limit}\limits_{x \to \infty } \,\frac{{{{\left( {{2^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}\,\, - \,\,{{\left( {{3^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}}}{{{x^n}}}\,$ (where $n \in N$) is

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