■5269 / inTopicNo.2) |
Re[1]
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□投稿者/ Kotatu 一般人(3回)-(2005/11/06(Sun) 18:18:41)
| f(x)=1/x∫(0〜π/2)sin(xy)dy =1/x[-cos(xy)/x](x=0〜π/2) =(1/x^2){-cos(πx/2)+1} =(1/x^2){1-cos(πx/2)} f(-6)=(1/6^2){1-cos(-6π/2)}=(1/6^2){1-cos(3π)} =(1/6^2){1+1}=1/18 f(x)=(1/x^2){sin(πx/2)}^2/{1+cos(πx/2)} =(1/x^2)(πx/2)^2(sint/t)^2/{1+cos(πx/2)} 、( t=πx/2 ) lim(x→0)f(x)=(π/2)^2・1^2/(1+1)=(π^2)/8
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