Question

分段函数f(x)={(1+x)^(1/x)-e,x≠0 0 ,x=0},求f（x）在点x=0处的导

Answer 1

f(x)

=(1+x)^(1/x)-e      , x≠0 

=0                          , x=0

lim(x->0) f(x) =lim(x->0) (1+x)^(1/x)-e =e-e =0 =f(0)  =>x=0, f(x) 连续

f'(0) 

= lim(h->0) [(1+h)^(1/h)-e -f(0) ]/h

=lim(h->0) [(1+h)^(1/h)-e]/h

=lim(h->0) { e^[ln(1+h)/h]-e }/h   

=lim(h->0) { e^[1-(1/2)h]-e }/h   

=lim(h->0)  e. { e^[-(1/2)h]-1 }/h   

=lim(h->0)  e. [-(1/2)h]/h   

=-(1/2)e