当x->0时无穷小α=x^2 与β=1-√(1-2x^2)的关系是
问题描述：
当x->0时无穷小α=x^2 与β=1-√(1-2x^2)的关系是
答案是α是β的等价无穷小这是怎么求出来的 ?求解释谢谢
答
lim (x->0)x^2/(1-√(1-2x^2))
=lim (x->0)x^2(1+√(1-2x^2))/[(1-√(1-2x^2))(1+√(1-2x^2))]
=lim (x->0)x^2(1+√(1-2x^2))/[(1-(1-2x^2))]
=lim (x->0)x^2(1+√(1-2x^2))/(2x^2)
=lim (x->0)(1+√(1-2x^2))/2
=(1+1)/2
=1
所以α是β的等价无穷小