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求n^2(e^(2+1/n)+e^(2-1/n)-2e^2)的极限,n趋于无穷,
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求n^2(e^(2+1/n)+e^(2-1/n)-2e^2)的极限,n趋于无穷,
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答案和解析
解法一:(泰勒公式法)
原式=lim(n->∞){e²n²[e^(1/n)+e^(-1/n)-2]}
=e²*lim(n->∞){n²[1+(1/n)+(1/n)²/2+1-(1/n)+(1/n)²/2+o(1/n³)-2]} (应用泰勒公式展开)
=e²*lim(n->∞){n²[1/n²+o(1/n³)]}
=e²*lim(n->∞)[1+o(1/n)]
=e²*1
=e²;
解法二:(洛必达法)
原式=lim(n->∞)lim(n->∞){e²n²[e^(1/n)+e^(-1/n)-2]}
=e²*lim(n->∞){n[e^(1/(2n))-e^(-1/(2n))]}²
=e²*lim(x->0){[e^(x/2)-e^(-x/2)]/x}² (令x=1/n)
=e²*{lim(x->0)[(e^(x/2)-e^(-x/2))/x]}²
=e²*{lim(x->0)[e^(x/2)/2+e^(-x/2)/2]}² (0/0型极限,应用罗比达法则)
=e²*(1/2+1/2)
=e².
原式=lim(n->∞){e²n²[e^(1/n)+e^(-1/n)-2]}
=e²*lim(n->∞){n²[1+(1/n)+(1/n)²/2+1-(1/n)+(1/n)²/2+o(1/n³)-2]} (应用泰勒公式展开)
=e²*lim(n->∞){n²[1/n²+o(1/n³)]}
=e²*lim(n->∞)[1+o(1/n)]
=e²*1
=e²;
解法二:(洛必达法)
原式=lim(n->∞)lim(n->∞){e²n²[e^(1/n)+e^(-1/n)-2]}
=e²*lim(n->∞){n[e^(1/(2n))-e^(-1/(2n))]}²
=e²*lim(x->0){[e^(x/2)-e^(-x/2)]/x}² (令x=1/n)
=e²*{lim(x->0)[(e^(x/2)-e^(-x/2))/x]}²
=e²*{lim(x->0)[e^(x/2)/2+e^(-x/2)/2]}² (0/0型极限,应用罗比达法则)
=e²*(1/2+1/2)
=e².
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