当x趋向多少时?y=x(x-1)(x+1)^(1/2)/(x^3-1)是无穷小

Question

当x趋向多少时?y=x(x-1)(x+1)^(1/2)/(x^3-1)是无穷小

Answer 1

y=x(x-1)(x+1)^(1/2)/(x^3-1)=x(x-1)(x+1)^(1/2)/[(x-1)(x^2+x+1)]
当x->1时,lim y=lim x(x-1)(x+1)^(1/2)/[(x-1)(x^2+x+1)]=lim x(x+1)^(1/2)/(x^2+x+1)=√2/3,不是无穷小.
当x≠1时,y=x(x+1)^(1/2)/(x^2+x+1),要想使y为无穷小,显然只需x->0或x->-1+.注意第二种情况是趋于-1+,而不是趋于-1或-1-.目的是为了保证(x+1)^(1/2)有意义.
不明白请追问.