如题所述
解:
∵lim(x->0)[(e^(x/2)-e^(-x/2))/x]=lim(x->0)[(e^(x/2)-e^(-x/2))'/x']
(0/0型极限,应用洛必达法则)
=lim(x->0)[(e^(x/2)+e^(-x/2))/2]=1
∴原式=lim(n->∞){n²*e²[e^(1/n)+e^(-1/n)-2]}
=lim(n->∞){n²*e²[e^(1/(2n))-e^(-1/(2n))]²} (因式分解)
=e²*lim(n->∞){[(e^(1/(2n))-e^(-1/(2n)))/(1/n)]²} (分子分母同除n²)
=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*1²
=e²。
∵lim(x->0)[(e^(x/2)-e^(-x/2))/x]=lim(x->0)[(e^(x/2)-e^(-x/2))'/x']
(0/0型极限,应用洛必达法则)
=lim(x->0)[(e^(x/2)+e^(-x/2))/2]=1
∴原式=lim(n->∞){n²*e²[e^(1/n)+e^(-1/n)-2]}
=lim(n->∞){n²*e²[e^(1/(2n))-e^(-1/(2n))]²} (因式分解)
=e²*lim(n->∞){[(e^(1/(2n))-e^(-1/(2n)))/(1/n)]²} (分子分母同除n²)
=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*1²
=e²。
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