Question

高数导数题

设f(0)>0,f'(0)存在,求lim(x->∞)[f(1/x)/f(0)]^x。

Answer 1

L=lim(x->∞) [f(1/x)/f(0)]^x
lnL
=lim(x->∞) [lnf(1/x)-lnf(0)] /(1/x) (0/0)
=lim(x->∞) [-(1/x^2)/f(1/x) ]/(-1/x^2)
=lim(x->∞) 1/f(1/x)
=1/f(0)
=>
L = e^[1/f(0)]

追问

您好，参考答案是 e^[f'(0)/f(0)],另外请问您在洛必达那步ln f(1/x) - ln f(0)是如何求导的，谢谢了

追答

不好意思
y= lnf(1/x)
y' = -(1/x^2). f'(1/x)/f(1/x)
/
L=lim(x->∞) [f(1/x)/f(0)]^x
lnL
=lim(x->∞) [lnf(1/x)-lnf(0)] /(1/x) (0/0)
=lim(x->∞) [-(1/x^2).f'(1/x)/f(1/x) ]/(-1/x^2)
=lim(x->∞) f'(1/x)/f(1/x)
=f'(0)/f(0)
=>
L = e^[f'(0)/f(0)]

Answer 2

y=6^x y'=6^x*ln6 所以： dy=6^x*ln6 dx