Question

高等数学:确定a,b,c的值，使下列极限等式成立。

如图

Answer 1

(1)

lim(x->+∞) [ √(x^2-x+1)-ax+b] =0

lim(x->+∞) [ (x^2-x+1)-(ax-b)^2]/[ √(x^2-x+1)+(ax-b)] =0

lim(x->+∞)  [(1-a^2)x^2+(-1+2ab)x +(1-b^2) ]/[ √(x^2-x+1)+(ax-b)] =0

分子分母同时除以x

lim(x->+∞)  [(1-a^2)x+(-1+2ab) +(1-b^2)/x ]/[ √(1-1/x+1/^2)+(a-b/x) ] =0

=>

1-a^2 =0                      (1) and 

-1+2ab=0                     (2) 

from (1)

a=1 or -1 (rej)

a=1

from (1)

-1+2b=0

b=1/2

ie ( a,b) = (1, 1/2)

(2)

x^2+3 = (x-1)^2 +2x+2 = (x-1)^2 +2(x-1) +4

let : y=x-1

x^2+3 = y^2+2y +4

y->0

√(x^2+3)

= √[ y^2+2y +4]

=2 √ [ 1+ (1/2)y + (1/4)y^2 ]

=2 . {  1+ (1/2)[(1/2)y + (1/4)y^2] -(1/8)[(1/2)y + (1/4)y^2]^2 +o(y^2) }

=2 . {  1+ (1/2)[(1/2)y + (1/4)y^2] -(1/8)[ (1/4)y^2+o(y^2)] +o(y^2) }

=2 . [  1+ (1/4)y + (3/32)y^2 +o(y^2) ]

= 2+ (1/2)y + (3/16)y^2 +o(y^2)

lim(x->1)  [a(x-1)^2+b(x-1) +c -√(x^2+3) ]/(x-1)^2 =0

lim(y->0)  [ ay^2+by +c + 2- (1/2)y - (3/16)y^2 +o(y^2)  ]/y^2 =0

lim(y->0)  [ (a -3/16 )y^2+(b- 1/2)y + (c - 2) +o(y^2)  ]/y^2 =0

=>

a -3/16  =0  and b- 1/2 =0 and c - 2=0

a=3/16  and b=1/2 and c=2

(a,b,c) = (3/16, 1/2, 2)

Answer 2

Answer 3

√（x²-x+1）-（ax-b）--》0
（x²-x+1）--》（ax-b）²=a²x²-2abx+b²
a²=1，-2ab=-1，
a=1，b=1/2
或者：
√（x²-x+1）=√[（x-1/2）²+3/4]--》√（x-1/2）²=x-1/2=ax-b
a=1，b=1/2