利用和差化积公式.lim [sin(√(n+1)-sin(√n)]=lim 2cos{[(√(n+1)+(√n)]/2}*sin{[(√(n+1)-(√n)]/2}x->∞ x->∞ =lim 2cos{[(√(n+1)+(√n)]/2}*sin{(1/2)/[(√(n+1)+(√n)]}x->∞ =lim 2cos{[(√(n+1)+(√n)]/2}...