We define ln(x)=int[(1 , x)(1/t)dt], then (ln(x))’=1/x . Define e^x as the inverse of ln(x) .
Then lim [ln(1+x)/x] =lim 1/(1+x) =1 as x approaches to 0 (using the Hp rule). Let ln(1+y)=x then y=e^x-1 and lim[ln(1+y)/y]=lim[x/e^x-1]=1 as x approaches to 0. then limit[(e^h-1)/h]=1 as h approaches to 0 . We have
lim[(5e^x-5e^(x+h))/3h]=(5/3)lim[(e^x-(e^x)(e^h))/h]=(5/3)e^xlim[(1-e^h))/h]=
-(5/3)e^xlim[(e^h-1))/h]=-(5/3)e^x as h approaches to 0.