作业帮设f(x)在[a,b]上连续,x1,x2,x3.xn∈[a,b],且t1+t2+t3+.+tn=1,ti>0,i=1,2,3...,n.证明：存在x0∈[a,b],使得f(x0)=t1f(x1) + t2f(x2) + .+ tnf(xn).利用归结原则证明：lim n→无穷 (1+1/n+1/n^2)^n=e. 在线等求解答.
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![设f(x)在[a,b]上连续,x1,x2,x3.xn∈[a,b],且t1+t2+t3+.+tn=1,ti>0,i=1,2,3...,n.证明：存在x0∈[a,b],使得f(x0)=t1f(x1) + t2f(x2) + .+ tnf(xn).利用归结原则证明：lim n→无穷 (1+1/n+1/n^2)^n=e. 在线等求解答.](/uploads/image/z/5580565-61-5.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2Cx1%2Cx2%2Cx3.xn%E2%88%88%5Ba%2Cb%5D%2C%E4%B8%94t1%2Bt2%2Bt3%2B.%2Btn%3D1%2Cti%26gt%3B0%2Ci%3D1%2C2%2C3...%2Cn.%E8%AF%81%E6%98%8E%EF%BC%9A%E5%AD%98%E5%9C%A8x0%E2%88%88%5Ba%2Cb%5D%2C%E4%BD%BF%E5%BE%97f%28x0%29%3Dt1f%28x1%29+%2B+t2f%28x2%29+%2B+.%2B+tnf%28xn%29.%E5%88%A9%E7%94%A8%E5%BD%92%E7%BB%93%E5%8E%9F%E5%88%99%E8%AF%81%E6%98%8E%EF%BC%9Alim+n%E2%86%92%E6%97%A0%E7%A9%B7+%281%2B1%2Fn%2B1%2Fn%5E2%29%5En%3De.+%E5%9C%A8%E7%BA%BF%E7%AD%89%E6%B1%82%E8%A7%A3%E7%AD%94.)
设f(x)在[a,b]上连续,x1,x2,x3.xn∈[a,b],且t1+t2+t3+.+tn=1,ti>0,i=1,2,3...,n.证明：存在x0∈[a,b],使得f(x0)=t1f(x1) + t2f(x2) + .+ tnf(xn).利用归结原则证明：lim n→无穷 (1+1/n+1/n^2)^n=e. 在线等求解答.
设f(x)在[a,b]上连续,x1,x2,x3.xn∈[a,b],且t1+t2+t3+.+tn=1,ti>0,i=1,2,3...,n.证明：存在x0∈[a,b],使得f(x0)=t1f(x1) + t2f(x2) + .+ tnf(xn).
利用归结原则证明：lim n→无穷 (1+1/n+1/n^2)^n=e. 在线等求解答.
设f(x)在[a,b]上连续,x1,x2,x3.xn∈[a,b],且t1+t2+t3+.+tn=1,ti>0,i=1,2,3...,n.证明：存在x0∈[a,b],使得f(x0)=t1f(x1) + t2f(x2) + .+ tnf(xn).利用归结原则证明：lim n→无穷 (1+1/n+1/n^2)^n=e. 在线等求解答.
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