Since [tex]x^2-4[/tex] is continuous at x=3 (basically no interruption of any form), we can do direct substitution and plug in x=3.
[tex]6 = \displaystyle\lim_{x\to3} f(x) = f(3) = \dfrac{a(3)^2-b}{(3)^2-4} = \dfrac{9a-b}{5}[/tex]
So then the relationship between a and b would be [tex]\boxed{9a-b = 30}[/tex]
Hope this helps.
