先看(x+1)^1/x的導數(shù)
令
f(x)=(x+1)^1/x
lnf(x)=ln(x+1)/x
兩端對x求導得
f'(x)/f(x)=[x/(x+1)-ln(x+1)]/x^2
f'(x)=[x/(x+1)-ln(x+1)]/x^2*(x+1)^1/x
lim(x→0) [(x+1)^1/x-e]/x (0/0)
=lim(x→0) [x/(x+1)-ln(x+1)]/x^2*(x+1)^1/x
=lim(x→0) [x/(x+1)-ln(x+1)]/x^2*lim(x→0) (x+1)^1/x
=e*lim(x→0) [x/(x+1)-ln(x+1)]/x^2
=e*lim(x→0) [x-(x+1)*ln(x+1)]/[x^2(x+1)]
=e*lim(x→0) [x-(x+1)*ln(x+1)]/x^2 (0/0)
=e*lim(x→0) [1-1-ln(x+1)]/(2x)
=e*lim(x→0) -ln(x+1)/(2x)
=e*lim(x→0) -x/(2x)
=e/2