F.4 M2 [limit]

2010-03-27 6:29 am
lim x->∞ {(x+4)[1+(2/x)-(15 / x^2)] / x} ^x

回答 (2)

2010-03-27 6:59 am
✔ 最佳答案


lim
x->∞_{(x+4)[1+(2/x)-(15/x^2)]/x}^x

=lim y->0+_{(1/y+4)[1+(2y)-(15y^2)]*y}^(1/y)

=lim y->0+_((1+4y)(1+2y-15y^2))^(1/y)

=e^lim y->0+_ln(((1+4y)(1+2y-15y^2))^(1/y))

=e^lim y->0+_(1/y)ln((1+4y)(1+2y-15y^2))

=e^lim y->0+_ln((1+4y)(1+2y-15y^2))/y

=e^lim y->0+_ln(-60y^3-6y^2+6y+1)/y
0/0 type

=e^lim y->0+_[(-180y^2-12y+6)/(-60y^3-6y^2+6y+1)]

=e^6



2010-03-27 5:34 pm
正如 3 = 3 + 1 - 1 加左1俾佢都要減番1
依家加左個ln , 一定要加返粒e先得 , 唔係既話你就"無中生有"左個ln
因為 3 = e^ ( ln 3)
因為 代y=0 入去果陣 分子,分母都 = 0 , 所以可以用 L hospital's rule ,
分子同分母可以同時對y微分一次 且 limit 不變。
http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule


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