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用定义证明n^a/c^n的极限为0(a>0,c>1)
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用定义证明n^a/c^n的极限为0(a>0,c>1)
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证明 lim(n→∞)(n^a)/(c^n) = 0(c>1,a>0) .
证 因 a>0,取 j=[a]+1,则 a≤j lim(n→∞)[(n^a)/(c^n)]
= lim(x→+∞)[(x^a)/(c^x)] (0/0)
= lim(x→+∞){[ax^(a-1)]/[(c^x)lnc]} (a-1>0,0/0)
= lim(x→+∞){[a(a-1)x^(a-2)]/[(c^x)(lnc)^2]} (a-2>0,0/0)
= ……
= lim(x→+∞){[a(a-1)…(a-j+1)x^(a-j)]/[(c^x)(lnc)^j]} (a-j≤0)
= lim(x→+∞){[a(a-1)…(a-j+1)]/[[x^(j-a)](c^x)(lnc)^j]} (a-j≤0)
= 0.
证 因 a>0,取 j=[a]+1,则 a≤j lim(n→∞)[(n^a)/(c^n)]
= lim(x→+∞)[(x^a)/(c^x)] (0/0)
= lim(x→+∞){[ax^(a-1)]/[(c^x)lnc]} (a-1>0,0/0)
= lim(x→+∞){[a(a-1)x^(a-2)]/[(c^x)(lnc)^2]} (a-2>0,0/0)
= ……
= lim(x→+∞){[a(a-1)…(a-j+1)x^(a-j)]/[(c^x)(lnc)^j]} (a-j≤0)
= lim(x→+∞){[a(a-1)…(a-j+1)]/[[x^(j-a)](c^x)(lnc)^j]} (a-j≤0)
= 0.
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