1.求极限lim [(1-1/2^2)(1-1/3^2)...(1-1/n^2)] n趋向于无穷大.

(2-1)(2+1)(3-1)(3+1).(n-1)(n+1)/2.3.4.5.n^2=(n+1)/2n=1/2+1/2n
x=pcosa y=psina a(0,90)p(0,3) ∫∫ln(1+x^2+y^2)dxdy=∫∫ln(1+p^2)pdadp
π/2∫ln(1+p^2)d1+p^2
第二步用的是分部积分法：
∵原式=π/2∫(0,3)pln(1+p²)dp (∫(0,1)表示从0到1积分)
=π/4∫(0,3)ln(1+p²)d(1+p²)
在分部积分公式中,设u=ln(1+p²),dv=d(1+p²)
则du=2pdp/(1+p²),v=1+p²
∴由分部积分公式∫udv=uv-∫vdu,得：
原式=π/4([(1+p²)ln(1+p²)]|(0,3)-2∫(0,3)pdp)
=π/4([(1+p²)ln(1+p²)-p²]|(0,3)
=π(10ln10-9)/4