三角函數的極限問題!

lim (x→∞) [sin√(x+1) - sin√x]
= 2 lim (x→∞) sin{[√(x+1) - √x]/2} cos √{[(x+1) + √x]/2}
= 2 lim (x→∞) sin{[√(x+1) - √x][√(x+1) + √x]/2[√(x+1) + √x]} cos √{[(x+1) + √x]/2}
= 2 lim (x→∞) sin{1/2[√(x+1) + √x]} cos √{[(x+1) + √x]/2}

觀察:

lim (x→∞) sin{1/2[√(x+1) + √x]}
= sin{lim (x→∞) 1/2[√(x+1) + √x]}
= sin 0
= 0

-1 ≦ cos √{[(x+1) + √x]/2} ≦ 1

利用壓迫定理,

2 lim (x→∞) sin{1/2[√(x+1) + √x]} cos √{[(x+1) + √x]/2} = 0

lim (x→∞) [sin√(x+1) - sin√x] = 0

2007-11-13 12:27:11 補充：
第十行出了亂碼, 應該是:

-1 <= cos √{[(x+1) + √x]/2} <= 1

sin((x+1)^0.5)-sin(x^0.5)
=2 sin(0.5*((x+1)^0.5-x^0.5)) * cos(0.5*((x+1)^0.5+x^0.5))

limx->inf [(x+1)^0.5-x^0.5] = 0
cos is a finite valued function

so limx→∞ (SIN(X+1)^1/2 - SINX^1/2 ) = 0