麻烦知道的大神告诉下,
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/02 00:43:19
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麻烦知道的大神告诉下,
![麻烦知道的大神告诉下,](/uploads/image/z/15624614-38-4.jpg?t=%E9%BA%BB%E7%83%A6%E7%9F%A5%E9%81%93%E7%9A%84%E5%A4%A7%E7%A5%9E%E5%91%8A%E8%AF%89%E4%B8%8B%2C)
1.证明xln[x+√(1+x²)]>√(1+x²)-1(x>0)
证明:设f(x)=xln[x+√(1+x²)]-√(1+x²)+1;由于当x>0时,
f '(x)=ln[x+√(1+x²)]+x[1+x/√(1+x²)]/[x+√(1+x²)]-x/√(1+x²)
=ln[x+√(1+x²)]+x/√(1+x²)-x/√(1+x²)
=ln[x+√(1+x²)]>0恒成立;即在x>0时,f(x)是增函数;又f(0)=0;故当x>0时恒有f(x)>0;
即有xln[x+√(1+x²)]>√(1+x²)-1(x>0).
2.求极限x→1lim[2/(x²-1)-3/(x³-1)]
原式=x→1lim[2/(x-1)(x+1)-3/(x-1)(x²+x+1)]
=x→1lim[2(x²+x+1)-3(x+1)]/[(x-1)(x+1)(x²+x+1)]
=x→1lim[(2x²-x-1)/(x-1)(x+1)(x²+x+1)]
=x→1lim[(2x+1)(x-1)/(x-1)(x+1)(x²+x+1)]
=x→1lim[(2x+1)/(x+1)(x²+x+1)]=1/2.
证明:设f(x)=xln[x+√(1+x²)]-√(1+x²)+1;由于当x>0时,
f '(x)=ln[x+√(1+x²)]+x[1+x/√(1+x²)]/[x+√(1+x²)]-x/√(1+x²)
=ln[x+√(1+x²)]+x/√(1+x²)-x/√(1+x²)
=ln[x+√(1+x²)]>0恒成立;即在x>0时,f(x)是增函数;又f(0)=0;故当x>0时恒有f(x)>0;
即有xln[x+√(1+x²)]>√(1+x²)-1(x>0).
2.求极限x→1lim[2/(x²-1)-3/(x³-1)]
原式=x→1lim[2/(x-1)(x+1)-3/(x-1)(x²+x+1)]
=x→1lim[2(x²+x+1)-3(x+1)]/[(x-1)(x+1)(x²+x+1)]
=x→1lim[(2x²-x-1)/(x-1)(x+1)(x²+x+1)]
=x→1lim[(2x+1)(x-1)/(x-1)(x+1)(x²+x+1)]
=x→1lim[(2x+1)/(x+1)(x²+x+1)]=1/2.