Hi, my question is this;

My strategy is to take the limit as x->1 of the first relation, which yields 2.

I then set the limit (x->1) ax^2 - bx + 5 = 2

or

a - b + 3 = 0 (1)

Likewise, I take the limit as x->3 of the second and third relations.

These two limits yield

9a - 3b + 5 = 12 - a + b or

10a -4b - 7 = 0 (2)

I now have two equations and two unknowns so I solve for a = -19/6 and b = -37/6.

What is wrong with this approach?

Thanks.