设斜渐近线为y=ax+b
则有
lim(x->+∞)Y/y=1
=lim(x->+∞)(2+x)^1.5/(x^0.5(ax+b)),(上下同时除以x^1.5)
=lim(x->+∞)(2/x+1)^1.5/(a+b/x))=1
则a=1
lim(x->+∞)Y-y=0
=lim(x->+∞)(2+x)^1.5/x^0.5-(ax+b),(分式上下同时除以x^1.5)
=lim(x->+∞)x(2/x+1)^1.5-(ax+b),等价无穷小展开
=lim(x->+∞)x(1+1.5*2/x)-(ax+b)
=lim(x->+∞)1.5*2-b=0
b=3
因此,斜渐近线为y=x+3