Question

高数 设函数y=f(x)由方程y-x=e^[x(1-y)]确定,求lim n->无穷 n[f(1/

高数 设函数y=f(x)由方程y-x=e^[x(1-y)]确定,求lim n->无穷 n[f(1/n)-1]=求具体推理过程 谢谢

Answer 1

7.f(0)=1,设u=1/n→0，对y-x=e^[x(1-y)]求导数得
y'-1=e^[x(1-y)]*(1-y-xy'),
∴{1+xe^[x(1-y)]}y'=1+(1-y)e^[x(1-y)],
∴y'(0)=1,
∴原式→[f(u)-1]/u→f'(0)=1.

追问

最后求出导数 怎么就变为极限为1了

追答

用罗比达法则。

追问

还是看不懂 我接着问吧

Answer 2

consider
lim(x->∞) x [ f(1/x) -1]
let
x=1/y

lim(x->∞) x [ f(1/x) -1]
=lim(y->0) [ f(y) -1] /y
=0

y-x= e^[x(1-y)]
y(0) =e^0 =1

y-x= e^[x(1-y)]
d/dx(y-x) = d/dx ( e^[x(1-y)] )
y' -1 = e^[x(1-y)] . ( 1 - xy' - y )
( 1+ xe^[x(1-y)] ) y' = (1-y).e^[x(1-y)]
y' =(1-y).e^[x(1-y)] /( 1+ xe^[x(1-y)] )
y'(0) =(1-y(0)).e^0
=0

f(y) = y(0)+ (y'(0)/1!)y+ (y''(0)/2!)y^2 +....
= 1 + (y''(0)/2!)y^2 +....
f(y)-1 =(y''(0)/2!)y^2 +.... order : at least y^2

追问

看不懂

追答

泰勒展式
f(x) = f(0) +[f'(0)/1!]x +[f'(0)/1!]x^2 +....
f(0)=1 , f'(0)=0

f(y) = 1 +0.y +[f''(0)/2!]y^2 +....
= 1 +[f''(0)/2!]y^2 +....
f(y) -1 =[f''(0)/2!]y^2 +....

分子 ： 阶 ： 最小 是2
分母 ： 阶 ：1
lim(y->0) [ f(y) -1] /y =0