sin(A+B)-sin(A-B)=2sinBcosA
A+B=√(x+1)
A-B=√x
A=(1/2)[√(x+1)+√x]
B=(1/2)[√(x+1)-√x]
|lim(x→+∞)(sin√(x+1)-sin√x)|
=|lim(x→+∞)2sin[(1/2)[√(x+1)-√x]]cos(1/2)[√(x+1)+√x] |<=0*1=0
所以lim(x→+∞)(sin√(x+1)-sin√x)=0
A+B=√(x+1)
A-B=√x
A=(1/2)[√(x+1)+√x]
B=(1/2)[√(x+1)-√x]
|lim(x→+∞)(sin√(x+1)-sin√x)|
=|lim(x→+∞)2sin[(1/2)[√(x+1)-√x]]cos(1/2)[√(x+1)+√x] |<=0*1=0
所以lim(x→+∞)(sin√(x+1)-sin√x)=0