求极限(1)lim(n->∞)∫(0,1)x^n/(1+x)dx (2)lim(n->∞)∫(n+k,n)sinx/xd
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求极限(1)lim(n->∞)∫(0,1)x^n/(1+x)dx (2)lim(n->∞)∫(n+k,n)sinx/xdx (k>0)
![求极限(1)lim(n->∞)∫(0,1)x^n/(1+x)dx (2)lim(n->∞)∫(n+k,n)sinx/xd](/uploads/image/z/226564-52-4.jpg?t=%E6%B1%82%E6%9E%81%E9%99%90%281%29lim%28n-%3E%E2%88%9E%29%E2%88%AB%280%2C1%29x%5En%2F%281%2Bx%29dx+%282%29lim%28n-%3E%E2%88%9E%29%E2%88%AB%28n%2Bk%2Cn%29sinx%2Fxd)
0 < xⁿ/(1 + x) < xⁿ
0 < ∫(0→1) xⁿ/(1 + x) dx < ∫(0→1) xⁿ dx = xⁿ⁺¹/(n + 1) |(0→1) = 1/(n + 1)
∵lim(n→∞) 1/(n + 1) = 0
∴lim(n→∞) ∫(0→1) xⁿ/(1 + x) dx = 0
0 ≤ |∫(n→n + k) (sinx)/x dx| ≤ ∫(n→n + k) |sinx|/|x| dx ≤ ∫(n→n + k) 1/n dx = k/n
∵lim(n→∞) k/n = 0
∴lim(n→∞) ∫(n→n + k) (sinx)/x dx = 0
再问: 1/n应为1/x吧
0 < ∫(0→1) xⁿ/(1 + x) dx < ∫(0→1) xⁿ dx = xⁿ⁺¹/(n + 1) |(0→1) = 1/(n + 1)
∵lim(n→∞) 1/(n + 1) = 0
∴lim(n→∞) ∫(0→1) xⁿ/(1 + x) dx = 0
0 ≤ |∫(n→n + k) (sinx)/x dx| ≤ ∫(n→n + k) |sinx|/|x| dx ≤ ∫(n→n + k) 1/n dx = k/n
∵lim(n→∞) k/n = 0
∴lim(n→∞) ∫(n→n + k) (sinx)/x dx = 0
再问: 1/n应为1/x吧
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