已知两曲线y=f(x)与y=∫e^(-t^)dt在点(0,0)处的切线相同,写出此切线方程,并求极限lim nf(2/n)

Question

如题所述

Answer 1

直接求y=∫e^(-t^)dt在点(0,0)处的导数，就是y'=e^[-(arctanx)²] /(1+x²) 则y'(0)=1
则显然切线方程是y=x
根据题意y=f(x)过点(0,0)。即f(0)=0

lim nf(2/n)= lim [f(2/n)-f(0)] / (1/n)
= 2 lim [f(2/n)-f(0)] / ( 2/n )
= 2 lim [f(2/n)-f(0)] / ( 2/n - 0 )
= 2·f'(0)
=2

Answer 2

y'=e^[-(arctanx)²] /(1+x²)         x=0,y'=e^[0]/1=1   

切线方程y=x

lim  nf(2/n)=   lim  2[f(2/n)-f(0)]/(2/n)

=2 f'(0)=2

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