# Find constants a and b such that lim x-->3 [sqrt(ax-b) - 1] / (x-3) = 4.?

Find constants a and b such that lim x-->3 [sqrt(ax-b) - 1] / (x-3) = 4.

I multiply both numerator and denominator by [sqrt(ax-b) + 1], and then notice that for the limit to exist ax-b + 1 must be a multiple of x-3, but after that I'm stuck.

Could someone please help me figure out how to do solve this limit?

Any input would be greatly appreciated!

### 2 Answers

- atsuoLv 61 month agoFavourite answer
Given : lim x→3 [√(ax - b) - 1] / (x - 3) = 4

[√(ax - b) - 1] / (x - 3)

= [√(ax - b) - 1][√(ax - b) + 1] / [(x - 3)[√(ax - b) + 1]]

= [ax - b - 1] / [(x - 3)[√(ax - b) + 1]] ---(#1)

When x approaches 3, the denominator approaches 0. But the fraction approaches 4, so the numerator must approach 0.

a(3) - b - 1 = 0

3a - 1 = b ---(#2)

Next, in (#1), the numerator must have the factor of (x - 3). So

ax - b - 1 = k(x - 3) ---(#3)

The coefficient of x must be equal in both sides of (#3), so k = a. Therefore

(#1) becomes

[ax - b - 1] / [(x - 3)[√(ax - b) + 1]]

= a(x - 3) / [(x - 3)[√(ax - b) + 1]]

= a / [√(ax - b) + 1] ---(#4)

When x = 3, (#4) must become (not approach) 4.

a / [√(3a - b) + 1] = 4

a/4 = √(3a - b) + 1

a = 4√(3a - b) + 4

a - 4 = 4√(3a - b)

a^2 - 8a + 16 = 16(3a - b)

a^2 - 56a + 16 = -16b

b = -(1/16)a^2 + (56/16)a - 1 ---(#5)

By (#2) and (#5),

3a - 1 = -(1/16)a^2 + (56/16)a - 1

48a - 16 = -a^2 + 56a - 16

a^2 - 8a = 0

a(a - 8) = 0

a = 0 or a = 8

If a = 0 then b = -1, so (#4) becomes 0/0 . This is an indeterminate form, so it can not become a solution.

If a = 8 then b = 23, so (#4) becomes 8/2 = 4 when x = 3. Therefore,

(a,b) = (8,23) is the solution.