作业帮lim(x→0)[(1+x²)^1/3-1]/1/3x² =lim(x→0)[1/3*(1+x²)^(-2/3)*2x]/[2x/3] 这步是怎么回事lim(x→0)[(1+x²)^1/3-1]/1/3x²=lim(x→0)[1/3*(1+x²)^(-2/3)*2x]/[2x/3] 这步是怎么回事?是这两个式子吗？ 忽略
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 21:26:03
![lim(x→0)[(1+x²)^1/3-1]/1/3x² =lim(x→0)[1/3*(1+x²)^(-2/3)*2x]/[2x/3] 这步是怎么回事lim(x→0)[(1+x²)^1/3-1]/1/3x²=lim(x→0)[1/3*(1+x²)^(-2/3)*2x]/[2x/3] 这步是怎么回事?是这两个式子吗？ 忽略](/uploads/image/z/5436654-6-4.jpg?t=lim%28x%E2%86%920%29%5B%281%2Bx%26%23178%3B%29%5E1%2F3-1%5D%2F1%2F3x%26%23178%3B+%3Dlim%28x%E2%86%920%29%5B1%2F3%2A%281%2Bx%26%23178%3B%29%5E%28-2%2F3%29%2A2x%5D%2F%5B2x%2F3%5D+%E8%BF%99%E6%AD%A5%E6%98%AF%E6%80%8E%E4%B9%88%E5%9B%9E%E4%BA%8Blim%28x%E2%86%920%29%5B%281%2Bx%26%23178%3B%29%5E1%2F3-1%5D%2F1%2F3x%26%23178%3B%3Dlim%28x%E2%86%920%29%5B1%2F3%2A%281%2Bx%26%23178%3B%29%5E%28-2%2F3%29%2A2x%5D%2F%5B2x%2F3%5D+%E8%BF%99%E6%AD%A5%E6%98%AF%E6%80%8E%E4%B9%88%E5%9B%9E%E4%BA%8B%3F%E6%98%AF%E8%BF%99%E4%B8%A4%E4%B8%AA%E5%BC%8F%E5%AD%90%E5%90%97%EF%BC%9F+++%E5%BF%BD%E7%95%A5)
lim(x→0)[(1+x²)^1/3-1]/1/3x² =lim(x→0)[1/3*(1+x²)^(-2/3)*2x]/[2x/3] 这步是怎么回事lim(x→0)[(1+x²)^1/3-1]/1/3x²=lim(x→0)[1/3*(1+x²)^(-2/3)*2x]/[2x/3] 这步是怎么回事?是这两个式子吗？ 忽略
lim(x→0)[(1+x²)^1/3-1]/1/3x² =lim(x→0)[1/3*(1+x²)^(-2/3)*2x]/[2x/3] 这步是怎么回事
lim(x→0)[(1+x²)^1/3-1]/1/3x²
=lim(x→0)[1/3*(1+x²)^(-2/3)*2x]/[2x/3] 这步是怎么回事?
是这两个式子吗？ 忽略图像
lim(x→0)[(1+x²)^1/3-1]/1/3x² =lim(x→0)[1/3*(1+x²)^(-2/3)*2x]/[2x/3] 这步是怎么回事lim(x→0)[(1+x²)^1/3-1]/1/3x²=lim(x→0)[1/3*(1+x²)^(-2/3)*2x]/[2x/3] 这步是怎么回事?是这两个式子吗？ 忽略
这是洛必达法则
分子分母分别求导
这是0/0式极限，用了洛必达法则
http://baike.baidu.com/image/b110e619b35df42bdab4bdbc