一道微积分求极限的题目lim(x→0)[(xcotx-1)/(x^2)]我的解法是(以下lim符号省略):原式=[(x/sinx)cosx-1]/(x^2)=[(1)cosx-1]/(x^2)为0/0形式,使用洛必达法则,得:(-sinx)/(2x)=-1/2但是却错了,答案是-1/3,请问上
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![一道微积分求极限的题目lim(x→0)[(xcotx-1)/(x^2)]我的解法是(以下lim符号省略):原式=[(x/sinx)cosx-1]/(x^2)=[(1)cosx-1]/(x^2)为0/0形式,使用洛必达法则,得:(-sinx)/(2x)=-1/2但是却错了,答案是-1/3,请问上](/uploads/image/z/7203057-33-7.jpg?t=%E4%B8%80%E9%81%93%E5%BE%AE%E7%A7%AF%E5%88%86%E6%B1%82%E6%9E%81%E9%99%90%E7%9A%84%E9%A2%98%E7%9B%AElim%28x%E2%86%920%29%5B%28xcotx-1%29%2F%28x%5E2%29%5D%E6%88%91%E7%9A%84%E8%A7%A3%E6%B3%95%E6%98%AF%EF%BC%88%E4%BB%A5%E4%B8%8Blim%E7%AC%A6%E5%8F%B7%E7%9C%81%E7%95%A5%EF%BC%89%EF%BC%9A%E5%8E%9F%E5%BC%8F%3D%5B%28x%2Fsinx%29cosx-1%5D%2F%28x%5E2%29%3D%5B%281%29cosx-1%5D%2F%28x%5E2%29%E4%B8%BA0%2F0%E5%BD%A2%E5%BC%8F%2C%E4%BD%BF%E7%94%A8%E6%B4%9B%E5%BF%85%E8%BE%BE%E6%B3%95%E5%88%99%2C%E5%BE%97%EF%BC%9A%28-sinx%29%2F%282x%29%3D-1%2F2%E4%BD%86%E6%98%AF%E5%8D%B4%E9%94%99%E4%BA%86%2C%E7%AD%94%E6%A1%88%E6%98%AF-1%2F3%2C%E8%AF%B7%E9%97%AE%E4%B8%8A)
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