作业帮求a,b.lim(x-->0)1/(bx-sinx)定积分[0-->x]t^2dt/(a+t)^(1/2)=1成立
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![求a,b.lim(x-->0)1/(bx-sinx)定积分[0-->x]t^2dt/(a+t)^(1/2)=1成立](/uploads/image/z/7110177-33-7.jpg?t=%E6%B1%82a%2Cb.lim%28x--%3E0%291%2F%28bx-sinx%29%E5%AE%9A%E7%A7%AF%E5%88%86%5B0--%3Ex%5Dt%5E2dt%2F%28a%2Bt%29%5E%281%2F2%29%3D1%E6%88%90%E7%AB%8B)
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求a,b.lim(x-->0)1/(bx-sinx)定积分[0-->x]t^2dt/(a+t)^(1/2)=1成立
求a,b.lim(x-->0)1/(bx-sinx)定积分[0-->x]t^2dt/(a+t)^(1/2)=1成立
求a,b.lim(x-->0)1/(bx-sinx)定积分[0-->x]t^2dt/(a+t)^(1/2)=1成立
应用洛必达法则原式=lim(x->0)x^2/√(a+x)*(b-cosx) 因x趋近于0,x^2趋近于0,而极限为1故b-cosx趋近于0,b=1 代入得lim(x->0)x^2/√