Question

请问高数极限怎么求

麻烦说下过程

Answer 1

第一个题目因式分解就可以，最后代入值，第二题用等价无穷小转化，

Answer 2

不好意思，第二个我不太确定，希望你能尽快解决。

Answer 3

5) 求极限部分：1/(1-x) - 3/(1-x^3) =[ (1+x+x^2)-3]/(1-x^3) = (x^2+x-2)/(1-x^3)=(x+2)(x-1) /(1-x^3) 

= -(x+2)/(x^2+x+1)。当x-->0时，极限＝-(1+2)/(1+1+1)=-1

13) (5-x^2)^(1/2) +x-3= [(5-x^2)^(1/2) +(x-3)]*  [(5-x^2)^(1/2) -(x-3)]/[(5-x^2)^(1/2) -(x-3)]

     = -2(x-1)(x-2) /[(5-x^2)^(1/2) -(x-3)]

   (x-1)^(1/2)-1= [(x-1)^(1/2)-1] *[(x-1)^(1/2)+1]/[(x-1)^(1/2)+1]=(x-2)/[(x-1)^(1/2)+1]

因此原求极限部分＝{-2(x-2) /[(5-x^2)^(1/2) -(x-3)]} /{(x-2)/{[(x-1)^(1/2)+1]}

=-2(x-1)[(x-1)^(1/2)+1]/[(5-x^2)^(1/2) -(x-3)]

当x-->2时，原极限＝2(2-1)[(2-1)^(1/2)+1]/[(5-2^2)^(1/2) -(2-3)]

＝2*2/ (1-2+3) = 2

Answer 4

Answer 5

第15题，因式分解
极限=[1/(1-x)-3/(1-x)/(1+x+x^2)]
=[(1+x+x^2)-3]/(1-x)/(1+x+x^2)
=[(x+x^2-2]/(1-x)/(1+x+x^2)
[(x-1)(x+2)]/(1-x)/(1+x+x^2)
=-(x+2)/(1+x+x^2)
=-3/3=-1