[再寄小读者之数学篇](2014-06-20 求极限---积分中值定理的应用)

证明: 当 $m<2$ 时, $\dps{\lim_{x\to 0^+}\cfrac{1}{x^m}\int_0^x \sin \cfrac{1}{t}\rd t=0}$.  

 

证明:

$$\beex \bea \lim_{x\to 0^+}\cfrac{1}{x^m}\int_0^x \sin \cfrac{1}{t}\rd t &=\lim_{x\to 0^+} \cfrac{1}{x^m} \int_0^x t^2\rd \cos \cfrac{1}{t}\\ &=\lim_{x\to 0^+} \cfrac{1}{x^m} \sex{x^2\cos\cfrac{1}{x} -\int_0^x 2t\cos\cfrac{1}{t}\rd t}\\ &=-2\lim_{x\to 0^+} \cfrac{\int_0^x t\cos \cfrac{1}{t}\rd t}{x^m}\quad\sex{2-m>0}\\ &=\lim_{x\to0^+}\cfrac{\sex{\xi_x\cos\cfrac{1}{\xi_x}}\cdot x}{x^m}\\ &=0. \eea \eeex$$