lim(sin(x+1)^1/2-sinx^1/2)x－无穷

lim(sin(x+1)^1/2-sinx^1/2)x－无穷
lim(sin(x+1)^1/2-sinx^1/2)
x－无穷

lim(sin(x+1)^1/2-sinx^1/2)x－无穷
用和差化积
sin(x+1)^1/2-sinx^1/2
=2cos{[√(x+1)+√x]/2}sin{[√(x+1)-√x]/2}
=2cos{[√(x+1)+√x]/2} sin{1/(2[√(x+1)+√x])}
因为x->∞,那么1/(2[√(x+1)+√x])->0,sin{1/(2[√(x+1)+√x])}->0
且cos{[√(x+1)+√x]/2}是有限值.
所以sin(x+1)^1/2-sinx^1/2=2cos{[√(x+1)+√x]/2} sin{1/(2[√(x+1)+√x])}->0
所以,原极限=0