高数一2.6求下列极限1) lim(x趋于∞)(1+k/x)^x (k≠0,整数)令y=k/x,则x=k/y ,x趋于∞等价于y趋于0,故lim(x趋于∞)(1+k/x)^x=lim(y趋于0)(1+y)^k/y=lim(y趋于0)[(1+y)^(1/y)]^k=[lim(y趋于0)[(1+y)^(1/y)]=e^k我
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![高数一2.6求下列极限1) lim(x趋于∞)(1+k/x)^x (k≠0,整数)令y=k/x,则x=k/y ,x趋于∞等价于y趋于0,故lim(x趋于∞)(1+k/x)^x=lim(y趋于0)(1+y)^k/y=lim(y趋于0)[(1+y)^(1/y)]^k=[lim(y趋于0)[(1+y)^(1/y)]=e^k我](/uploads/image/z/4055520-48-0.jpg?t=%E9%AB%98%E6%95%B0%E4%B8%802.6%E6%B1%82%E4%B8%8B%E5%88%97%E6%9E%81%E9%99%901%EF%BC%89+lim%28x%E8%B6%8B%E4%BA%8E%E2%88%9E%29%281%2Bk%2Fx%29%5Ex+%EF%BC%88k%E2%89%A00%2C%E6%95%B4%E6%95%B0%EF%BC%89%E4%BB%A4y%3Dk%2Fx%2C%E5%88%99x%3Dk%2Fy+%2Cx%E8%B6%8B%E4%BA%8E%E2%88%9E%E7%AD%89%E4%BB%B7%E4%BA%8Ey%E8%B6%8B%E4%BA%8E0%2C%E6%95%85lim%28x%E8%B6%8B%E4%BA%8E%E2%88%9E%EF%BC%89%EF%BC%881%2Bk%2Fx%29%5Ex%3Dlim%28y%E8%B6%8B%E4%BA%8E0%EF%BC%89%281%2By%29%5Ek%2Fy%3Dlim%28y%E8%B6%8B%E4%BA%8E0%EF%BC%89%5B%EF%BC%881%2By%29%5E%281%2Fy%29%5D%5Ek%3D%5Blim%28y%E8%B6%8B%E4%BA%8E0%EF%BC%89%5B%EF%BC%881%2By%29%5E%281%2Fy%29%5D%3De%5Ek%E6%88%91)
高数一2.6求下列极限1) lim(x趋于∞)(1+k/x)^x (k≠0,整数)令y=k/x,则x=k/y ,x趋于∞等价于y趋于0,故lim(x趋于∞)(1+k/x)^x=lim(y趋于0)(1+y)^k/y=lim(y趋于0)[(1+y)^(1/y)]^k=[lim(y趋于0)[(1+y)^(1/y)]=e^k我
高数一2.6求下列极限
1) lim(x趋于∞)(1+k/x)^x (k≠0,整数)
令y=k/x,则x=k/y ,x趋于∞等价于y趋于0,故
lim(x趋于∞)(1+k/x)^x=lim(y趋于0)(1+y)^k/y
=lim(y趋于0)[(1+y)^(1/y)]^k
=[lim(y趋于0)[(1+y)^(1/y)]
=e^k
我不懂的是为什么
lim(y趋于0)[(1+y)^(1/y)]^k=[lim(y趋于0)(1+y)^(1/y)]^y
等式中的k次方可以移到lim括号左边去这是为什么?
高数一2.6求下列极限1) lim(x趋于∞)(1+k/x)^x (k≠0,整数)令y=k/x,则x=k/y ,x趋于∞等价于y趋于0,故lim(x趋于∞)(1+k/x)^x=lim(y趋于0)(1+y)^k/y=lim(y趋于0)[(1+y)^(1/y)]^k=[lim(y趋于0)[(1+y)^(1/y)]=e^k我
lim(y趋于0)[(1+y)^(1/y)]^k=[lim(y趋于0)(1+y)^(1/y)]^K
k是常数啊,
0.9循环,趋向于1吧,你0.9循环的10次和1的10次是一样的吧
不是很明白你想问什么
k当是常数,可以移出括号外,而limy->0 (1+y)^(1/y)=e
所以limy->0 [(1+y)^(1/y)]^k=e^k