lim(n->∞) n^2 [x^(1/n)-x^(1/n+1)]答案是lnx没弄错。原题是x开n次方 x开n+1次方,没看懂
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lim(n->∞)n^2[x^(1/n)-x^(1/n+1)]答案是lnx没弄错。原题是x开n次方x开n+1次方,没看懂lim(n->∞)n^2[x^(1/n)-x^(1/n+1)]答案是lnx没弄错
lim(n->∞) n^2 [x^(1/n)-x^(1/n+1)]答案是lnx没弄错。原题是x开n次方 x开n+1次方,没看懂
lim(n->∞) n^2 [x^(1/n)-x^(1/n+1)]
答案是lnx
没弄错。原题是x开n次方 x开n+1次方,
没看懂
lim(n->∞) n^2 [x^(1/n)-x^(1/n+1)]答案是lnx没弄错。原题是x开n次方 x开n+1次方,没看懂
哈哈
把x看成常量
1/n看作x1,1/(n+1)看作x2
n^2 [x^(1/n)-x^(1/n+1)]=[(n+1)/n][1/n-1/(n+1)]*[x^(1/n)-x^(1/n+1)]
[1/n-1/(n+1)]*[x^(1/n)-x^(1/n+1)]为a^x在0处的导数
即得答案是lnx
确定没弄错题目?~~~~~~能看懂他写的吗?他是利用导数的定义做的,我快吃饭了,吃完再解释~~~~~~
解释不了! 感觉不对,另请高手吧!
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