设f(x)在[a,b]上连续,x1,x2,x3.xn∈[a,b],且t1+t2+t3+.+tn=1,ti>0,i=

2.令f(x)=(1+x+x²)^(1/x),则
lim{x→0}f(x)=lim{x→0}[(1+x+x²)^(1/x)]
=lim{x→0}e^[1/x*ln(1+x+x²)]
=e^[lim{x→0}1/x*ln(1+x+x²)]
=e^[lim{x→0}1/x*(x+x²)] 当a→0时,ln(1+a)~a
=e^[lim{x→0}(1+x)]
=e
取x{n}=1/n,则lim{n→∞}x{n}=lim{n→∞}1/n=0
由归结原则可知,lim{n→∞}f[x{n}]=lim{x→0}f(x)
即lim{n→∞}(1+1/n+1/n²)^n=e