f(x)=两种情况,f(x)=(1+x)^(1/x),x不等于0时；f(x)=k,x=0时,且f(x)在x=0点连续,求f'(x)

Question

f(x)=两种情况,f(x)=(1+x)^(1/x),x不等于0时；f(x)=k,x=0时,且f(x)在x=0点连续,求f'(x)

Answer 1

f(x)在x=0点连续
lim(x→0) f(x)
=lim(x→0) (1+x)^(1/x)
=e
所以k=e
现在看(1+x)^(1/x)的导数
[(1+x)^(1/x)]'={e^[ln(1+x)^(1/x)]}'
=e^[ln(1+x)^(1/x)]*[ln(1+x)^(1/x)]'
=(1+x)^(1/x)*[ln(1+x)/x)]'
=(1+x)^(1/x)*[x/(1+x)-ln(1+x)]/x^2
=(1+x)^(1/x)*[x-(1+x)ln(1+x)]/[(1+x)x^2]
f'(x)=lim(x→0) [f(x)-f(0)]/x
=lim(x→0) [(1+x)^(1/x)-e]/x (0/0)
=lim(x→0) (1+x)^(1/x)*[x-(1+x)ln(1+x)]/[(1+x)x^2]
=lim(x→0) (1+x)^(1/x)*lim(x→0) [x-(1+x)ln(1+x)]/[(1+x)x^2]
=e*lim(x→0) [x-(1+x)ln(1+x)]/[(1+x)x^2]
=e*lim(x→0) [x-(1+x)ln(1+x)]/x^2 (0/0)
=e*lim(x→0) [1-ln(1+x)-1]/(2x)
=e*lim(x→0) -ln(1+x)/(2x)
=e*lim(x→0) -x/(2x)
=-e/2