极限题:lim[n*tan(1/n)]^(n^2) (n趋于无穷)
来源:百度知道 编辑:UC知道 时间:2024/06/17 20:57:30
要有具体过程~
把数列的极限化成函数的极限
lim{[n*tan(1/n)]^(n^2),n→∞}
=lim[(tanx/x)^(1/x^2),x→0]
函数式取对数后的极限
lim{ln[(tanx/x)^(1/x^2)],x→0}
=lim[(lntanx-lnx)/x^2,x→0]
=lim[(2/sin2x-1/x)/(2x),x→0]
=lim[(x-sin2x/2)/(x^2sin2x),x→0]
=lim[2x/(sin2x),x→0]lim[(x-sin2x/2)/(2x^3),x→0]
=1*lim[(1-cos2x)/(6x^2),x→0]
=lim[(2(sinx)^2)/(6x^2),x→0]
=1/3
所以
lim[(tanx/x)^(1/x^2),x→0]
=lim{e^ln[(tanx/x)^(1/x^2)],x→0]
=e^(1/3)
即
lim{[n*tan(1/n)]^(n^2),n→∞}=e^(1/3)
先求limln[n*tan(1/n)]^(n^2)
=lim(n->∞)(n^2)ln[n*tan(1/n)](令1/n = x,x->0)
=lim(x->0)[ln(tanx/x)]/x^2
=lim(x->0)(x-sinxcosx)/2x^3(洛必达,sin2x~ 2x)
=lim(x->0)(1-cos2x)/6x^2
=lim(x->0) 1/2*(2x)^2/6x^2
=lim(x->0)2x^2/6x^2
=1/3
所以lim(n->∞)[n*tan(1/n)]^(n^2)=e^(1/3)
111
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