数分的三个求极限和一个证明题
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数分的三个求极限和一个证明题
![](http://img.wesiedu.com/upload/5/60/560b66692b89295cced0ea8c2140072b.jpg)
只要做2.证明题就行了。
![](http://img.wesiedu.com/upload/5/60/560b66692b89295cced0ea8c2140072b.jpg)
只要做2.证明题就行了。
![数分的三个求极限和一个证明题](/uploads/image/z/13387914-18-4.jpg?t=%E6%95%B0%E5%88%86%E7%9A%84%E4%B8%89%E4%B8%AA%E6%B1%82%E6%9E%81%E9%99%90%E5%92%8C%E4%B8%80%E4%B8%AA%E8%AF%81%E6%98%8E%E9%A2%98)
证:lim(x→0+)x^f(x)
=lim(x→0+)e^(f(x)lnx)
=e^(lim(x→0+)f(x)lnx)
∵f′+(0)存在,f(0)=0,
∴f′+(0)=lim(x→0+)(f(x)-f(0))/(x-0)=lim(x→0+)f(x)/x存在,
∴f(x)与x是同阶无穷小,
∵lim(x→0+)xlnx
=lim(x→0+)lnx/(1/x)
=lim(x→0+)(1/x)/(-1/x²)
=lim(x→0+)-x=0,
∴lim(x→0+)x/(1/lnx)=0,
∴1/lnx是比x低价的无穷小,
∴1/lnx是比f(x)低价的无穷小,
∴lim(x→0+)f(x)lnx
=lim(x→0+)f(x)/(1/lnx)=0,
∴lim(x→0+)x^f(x)
=e^(lim(x→0+)f(x)lnx)
=e^0=1.
=lim(x→0+)e^(f(x)lnx)
=e^(lim(x→0+)f(x)lnx)
∵f′+(0)存在,f(0)=0,
∴f′+(0)=lim(x→0+)(f(x)-f(0))/(x-0)=lim(x→0+)f(x)/x存在,
∴f(x)与x是同阶无穷小,
∵lim(x→0+)xlnx
=lim(x→0+)lnx/(1/x)
=lim(x→0+)(1/x)/(-1/x²)
=lim(x→0+)-x=0,
∴lim(x→0+)x/(1/lnx)=0,
∴1/lnx是比x低价的无穷小,
∴1/lnx是比f(x)低价的无穷小,
∴lim(x→0+)f(x)lnx
=lim(x→0+)f(x)/(1/lnx)=0,
∴lim(x→0+)x^f(x)
=e^(lim(x→0+)f(x)lnx)
=e^0=1.