题目
lim n → ∞ n ∫ 0 1 x n 1 + x d x \mathop {\lim }\limits_{n \to \infty } n\int_0^1 {{{{x^n}} \over {1 + x}}dx} n→∞limn∫011+xxndx
来源
证明
lim n → ∞ n ∫ 0 1 x n 1 + x d x = lim n → ∞ n n + 1 ∫ 0 1 1 1 + x d x n + 1 = 1 2 + lim n → ∞ ∫ 0 1 x n + 1 ( 1 + x ) 2 d x \mathop {\lim }\limits_{n \to \infty } n\int_0^1 {{{{x^n}} \over {1 + x}}dx} = \mathop {\lim }\limits_{n \to \infty } {n \over {n + 1}}\int_0^1 {{1 \over {1 + x}}d{x^{n + 1}}} = {1 \over 2} + \mathop {\lim }\limits_{n \to \infty } \int_0^1 {{{{x^{n + 1}}} \over {{{\left( {1 + x} \right)}^2}}}dx} n→∞limn∫011+xxndx=n→∞limn+1n∫011+x1dxn+1=21+n→∞lim∫01(1+x)2xn+1dx
法一： 0 < ∫ 0 1 x n + 1 ( 1 + x ) 2 d x < ∫ 0 1 x n + 1 d x 0 < \int_0^1 {{{{x^{n + 1}}} \over {{{\left( {1 + x} \right)}^2}}}dx} < \int_0^1 {{x^{n + 1}}dx} 0<∫01(1+x)2xn+1dx<∫01xn+1dx且
lim n → ∞ ∫ 0 1 x n + 1 d x = 0 \mathop {\lim }\limits_{n \to \infty } \int_0^1 {{x^{n + 1}}dx} = 0 n→∞lim∫01xn+1dx=0
根据夹逼定理可知 lim n → ∞ ∫ 0 1 x n + 1 ( 1 + x ) 2 d x = 0. \mathop {\lim }\limits_{n \to \infty } \int_0^1 {{{{x^{n + 1}}} \over {{{\left( {1 + x} \right)}^2}}}dx}=0. n→∞lim∫01(1+x)2xn+1dx=0.则 lim n → ∞ n ∫ 0 1 x n 1 + x d x = 1 2 \mathop {\lim }\limits_{n \to \infty } n\int_0^1 {{{{x^n}} \over {1 + x}}dx} = {1 \over 2} n→∞limn∫011+xxndx=21
法二：根据积分中值定理可知 lim n → ∞ ∫ 0 1 x n + 1 ( 1 + x ) 2 d x = lim n → ∞ 1 ( 1 + ξ ) 2 ∫ 0 1 x n + 1 d x = 0 \mathop {\lim }\limits_{n \to \infty } \int_0^1 {{{{x^{n + 1}}} \over {{{\left( {1 + x} \right)}^2}}}dx} = \mathop {\lim }\limits_{n \to \infty } {1 \over {{{\left( {1 + \xi } \right)}^2}}}\int_0^1 {{x^{n + 1}}dx} = 0 n→∞lim∫01(1+x)2xn+1dx=n→∞lim(1+ξ)21∫01xn+1dx=0其中 ξ \xi ξ介于 0 0 0与 1 1 1之间.
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