高数习题11.2

1. 求下列函数的极限：
   (1) ${\lim }_{k\to 0}{\int }_0^{\frac{\pi }{2}}\frac{d\phi }{\sqrt{1-k^{2}si{n}^2\phi }}$
   (2) ${\lim }_{k\to 1-0}{\int }_0^{\frac{\pi }{2}}\sqrt{1-k^{2}si{n}^2\phi }d\phi$
   (3) ${\lim }_{\alpha \to 0}{\int }_{\alpha }^{\alpha +\frac{\pi }{2}}\frac{si{n}^{2}x}{4+\alpha x^2}dx$
   (4) ${\lim }_{\alpha \to 0}{\int }_0^{1+\alpha }\frac{dx}{1+x^2+{\alpha }^2}$
   (5) ${\lim }_{y\to 0}{\int }_0^1\frac{e^{x}sinxy}{y+1}dx$
   解：
   (1) 二元函数$\frac{1}{\sqrt{1-k^{2}si{n}^2\phi }}$在$[0,\frac{\pi }{2}]\times [-\frac{1}{2},\frac{1}{2}]$上连续，所以${\lim }_{k\to 0}{\int }_0^{\frac{\pi }{2}}\frac{d\phi }{\sqrt{1-k^{2}si{n}^2\phi }}={\int }_0^{\frac{\pi }{2}}{\lim }_{k\to 0}\frac{d\phi }{\sqrt{1-k^{2}si{n}^2\phi }}=\frac{\pi }{2}$
   (2) 二元函数$\sqrt{1-k^{2}si{n}^2\phi }$在$[0,\frac{\pi }{2}]\times [0,1]$上连续，所以${\lim }_{k\to 1-0}{\int }_0^{\frac{\pi }{2}}\sqrt{1-k^{2}si{n}^2\phi }d\phi ={\int }_0^{\frac{\pi }{2}}{\lim }_{k\to 1-0}\sqrt{1-k^{2}si{n}^2\phi }d\phi ={\int }_0^{\frac{\pi }{2}}cos\phi d\phi =1$
   (3) 二元函数$\frac{si{n}^{2}x}{4+\alpha x^2}$在$[\alpha ,\alpha +\frac{\pi }{2}]\times [-1,1]$上连续，所以${\lim }_{\alpha \to 0}{\int }_{\alpha }^{\alpha +\frac{\pi }{2}}$