lim(1+1/x^2)^x x趋向于无穷大的极限
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lim(1+1/x^2)^xx趋向于无穷大的极限lim(1+1/x^2)^xx趋向于无穷大的极限lim(1+1/x^2)^xx趋向于无穷大的极限lim(x→∞)【(1+1/x²)^x】=li
lim(1+1/x^2)^x x趋向于无穷大的极限
lim(1+1/x^2)^x x趋向于无穷大的极限
lim(1+1/x^2)^x x趋向于无穷大的极限
lim(x→∞)【(1+1/x²)^x】
=lim(x→∞)【(1+1/x²)^x²▪1/x】
=lim(x→∞){【(1+1/x²)^x²】^1/x}
=lim(x→∞)【(1+1/x²)^x²】^lim(x→∞)(1/x)
=eº
=1
希望可以帮到你,有不懂的欢迎追问哦!
极限是1
简单说一下方法吧
(1+1/x^2)^x = e^(xln(1+1/x^2))
设x=1/t 带入,使用洛必达法则,得出结论
没算错的话是1谢谢采纳
e=lim n->∞ (1+1/n)^n
lim(1+1/x^2)^x
=lim((1+1/x^2)^x^2)/x
=lim e^(1/x)
=e^0
=1
利用x趋向于无穷大时,(1+ 1/x)^x=e. x趋向于无穷大时, (1+1/x^2)^x=(1+1/x^2)^[(x^2)/x] =e^0=1
1
ans =
1
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