math question?????????????????????

Question

find the values for the constants a and b such that
limit as x approaches o
the numerator is square root (a + bx) -square root of 3
and the denominator is x
and the function is equal to the square root of 3
so u have to find what a b are
any questions email me at [e-mail]
if u want to see the equation

Lim x→0 (√(a + bx) - √3) / x = √3
find a and b
thx
show the steps

Answer 1

Lim x→0 (√(a + bx) - √3) / x = √3

f(x) = (√(a + bx) - √3) / x
= (√a√(1 + (b/a)x) - √3) / x

This only becomes indeterminate if a=3 , hence we can't use de l'Hospital.
It's all coming back to me now... when you have non-integer powers you use a Binomial Expansion:
(a + bx)^1/2 ≈ √a + 1/2√(b/a) x +...

f(x) ≈ (√a + 1/2√(b/a) x - √3) / x

I can see the obvious solution but don't know how to prove it's unique:
a=3, then √a-√3 cancels out.

Then f(x) ≈ ( 1/2√(b/a) x ) / x = √(b/3) / 2

√(b/3) / 2 = √3
=> (b/3) / 4 = 3
b = 36

SOLUTION: a=3, b=36

and just for curiosity, f(x) reduces to:
f(x) = (√3√(1 + (36/3)x) - √3) / x
= √3 (√(1 + 12x) - 1) / x