解:
属于∞/∞型,应用
罗比塔法则分子
分母同时求导,得到极限:
= lim(x→∞)[√(x+2)-√3]’/(x-1)’
= lim(x→∞){1/[2√(x+2)]-0}/1
= lim(x→∞)1/[2√(x+2)]
=0
或者分子有理化为:
lim(x→∞)√(x+2)-√3/(x-1)
= lim(x→∞)[√(x+2)-√3][ √(x+2)+√3]/{(x-1)[√(x+2)+√3]}
= lim(x→∞)(x-1)*/{(x-1)[√(x+2)+√3]}
= lim(x→∞)1/[√(x+2)+√3]
=0