(a^Δx-1)/Δx=β/loga(1+β)=1/loga[(1+β)^(1/β)]
β→0时,lim ln[(1+β)^1/β]=lim [ln(1+β)/β]=lim [ln(1+β)]'/(β)'=lim1/(1+β)=1,
所以β→0时,lim [(1+β)^(1/β)]=e,
其实如果可用(ln x)'=1/x,则y=a^x,lny=xlna,(lny)'=y'/y=(xlna)'=lna,所以y'=ylna=lna*a^x
(a^Δx-1)/Δx=β/loga(1+β)=1/loga[(1+β)^(1/β)]
β→0时,lim ln[(1+β)^1/β]=lim [ln(1+β)/β]=lim [ln(1+β)]'/(β)'=lim1/(1+β)=1,
所以β→0时,lim [(1+β)^(1/β)]=e,
其实如果可用(ln x)'=1/x,则y=a^x,lny=xlna,(lny)'=y'/y=(xlna)'=lna,所以y'=ylna=lna*a^x