本文主要是介绍考研数学精选题目015，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目
lim n → ∞ sin [ π ln ( 2 n + 1 ) − ln ( 2 n ) ] \mathop {\lim }\limits_{n \to \infty } \sin \left[ {{\pi \over {\ln \left( {2n + 1} \right) - \ln \left( {2n} \right)}}} \right] n→∞limsin[ln(2n+1)−ln(2n)π]
来源
证明
lim n → ∞ sin [ π ln ( 2 n + 1 ) − ln ( 2 n ) ] = lim n → ∞ sin [ π ln ( 2 n + 1 ) − ln ( 2 n ) − 2 n π ] = lim n → ∞ sin π [ 1 − 2 n ln ( 1 + 1 2 n ) ln ( 1 + 1 2 n ) ] = lim n → ∞ sin π [ 2 n ( 1 2 n − ln ( 1 + 1 2 n ) ) ln ( 1 + 1 2 n ) ] = lim n → ∞ sin π 1 2 = 1 \mathop {\lim }\limits_{n \to \infty } \sin \left[ {{\pi \over {\ln \left( {2n + 1} \right) - \ln \left( {2n} \right)}}} \right] = \mathop {\lim }\limits_{n \to \infty } \sin \left[ {{\pi \over {\ln \left( {2n + 1} \right) - \ln \left( {2n} \right)}} - 2n\pi } \right] = \mathop {\lim }\limits_{n \to \infty } \sin \pi \left[ {{{1 - 2n\ln \left( {1 + {1 \over {2n}}} \right)} \over {\ln \left( {1 + {1 \over {2n}}} \right)}}} \right] = \mathop {\lim }\limits_{n \to \infty } \sin \pi \left[ {{{2n\left( {{1 \over {2n}} - \ln \left( {1 + {1 \over {2n}}} \right)} \right)} \over {\ln \left( {1 + {1 \over {2n}}} \right)}}} \right] = \mathop {\lim }\limits_{n \to \infty } \sin \pi {1 \over 2} = 1 n→∞limsin[ln(2n+1)−ln(2n)π]=n→∞limsin[ln(2n+1)−ln(2n)π−2nπ]=n→∞limsinπ[ln(1+2n1)1−2nln(1+2n1)]=n→∞limsinπ[ln(1+2n1)2n(2n1−ln(1+2n1))]=n→∞limsinπ21=1其中 1 2 n − ln ( 1 + 1 2 n ) ∼ 1 4 n 2 , n → ∞ {1 \over {2n}} - \ln \left( {1 + {1 \over {2n}}} \right)\sim{1 \over {4{n^2}}},n \to \infty 2n1−ln(1+2n1)∼4n21,n→∞
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