4个回答
展开全部
把n²看成1/(1/n²),然后用洛必达法则,得
原式=-1/2lim[-2sin(θ/n)cos(θ/n)*θ/n²]/[-2/n³]
=-1/2limθ/2*sin(2θ/n)/(1/n) (再用洛必达法则)
=-1/2limθ/2*[cos(2θ/n)*(-2θ/n²)]/(-1/n²)
=-1/2limθ/2*2θcos(2θ/n)
=-1/2*θ/2*2θlimcos(2θ/n)
=-θ²/2*limcos(2θ/n)
=-θ²/2*cos0
=-θ²/2
原式=-1/2lim[-2sin(θ/n)cos(θ/n)*θ/n²]/[-2/n³]
=-1/2limθ/2*sin(2θ/n)/(1/n) (再用洛必达法则)
=-1/2limθ/2*[cos(2θ/n)*(-2θ/n²)]/(-1/n²)
=-1/2limθ/2*2θcos(2θ/n)
=-1/2*θ/2*2θlimcos(2θ/n)
=-θ²/2*limcos(2θ/n)
=-θ²/2*cos0
=-θ²/2
展开全部
-(1/2)lim(n→∞) n²sin²(θ/n)
=-(1/2)lim(k→0) sin²(kθ)/k² ........ 倒代换k=1/n
=-(1/2)lim(k→0) (kθ)²/k² ........等价无穷小替换sin(kθ)~kθ
=-θ²/2
=-(1/2)lim(k→0) sin²(kθ)/k² ........ 倒代换k=1/n
=-(1/2)lim(k→0) (kθ)²/k² ........等价无穷小替换sin(kθ)~kθ
=-θ²/2
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展开全部
只是用了个等价无穷小:sinx ~x(x-->0)
lim n^2* [sin(θ/n)]^2
= lim n^2* (θ/n)^2
= θ^2
lim n^2* [sin(θ/n)]^2
= lim n^2* (θ/n)^2
= θ^2
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