Finding derivative given $f(x)$ and limit this is my very first post here. I'm a longtime lurker, I have learned a lot from this site and I hope I can, even though I'm still a student, be of any help to others. 
I have been working on the following exercise. Given:
$ f(4) = a, f'(4) = b
$ and 
$$ \lim_{x\to 2} = {f(3x-2)-5 \over (x^2-4)} = 9$$
Find a and b.
I already found the value of $a$ following this reasoning: $f(4)$ must be such a value that, when replaced in the given limit, yields zero; otherwise, the limit won't be $9$ but infinity. Under the assumption that I can factor the denominator easily -and simplify $(x-2)$ from $f(x)$ I found $a=f(4)=5$ and it coincides with the answer. I have, however, no reasonable clue of how to approach the $f'(4)=b$ part. All the information I seem to have is that the derivative is continuous, and that $(x-2)$ must be a factor of $f(x)$ .If it's derivative is continuous over Reals, then I infer that it is not a rational function. But I still cant see the right path towards the solution.
I'd be extremely grateful if someone could give me a hint, so I can work it out by myself
Thanks in advance!
 A: Since $x^2-4 \to 0$ and $f(3x-2)-5 \to 0$ (due to the continuity of $f$ and the your conclusion that $f(4) = 5)$ as $x \to 2$, L'Hospital's rule is applicable, and we get
$$ \lim_{x\to 2} \frac{f(3x-2)-5}{x^2-4} = \lim_{x\to 2} \frac{3f'(3x-2)}{2x} = \frac{3}{4}f'(4). $$
EDIT: If you don't want to use L'Hospital's rule, then you can also notice that
$$ \lim_{x\to 2} \frac{f(3x-2)-5}{x^2-4} = \lim_{x\to 2} \frac{1}{x+2} \cdot\lim_{x\to 2} \frac{f(3x-2)-5}{x-2} = \frac{1}{4} \lim_{x\to 2} \frac{f(3x-2)-f(4)}{x-2}. $$
The latter limit is by definition the derivative of the function $g(x) = f(3x-2)$ at $x=2$, which, by the chain rule, is
$$ g'(2) = 3f'(4). $$
A: Let $t = 3x-2$ so $x = \frac{t+2}3$.
Then
$$9 = \lim_{x\to 2}\frac{f(3x-2)-5}{x^2-4} = \lim_{x\to 2}\frac{f(3x-2)-f(4)}{(x-2)(x+2)} = \frac14 \lim_{x\to 2}\frac{f(3x-2)-f(4)}{x-2} = \frac14 \lim_{t \to 4}\frac{f(t)-f(4)}{\frac{t-4}3} = \frac34 f'(4)$$
A: If $3x-2=y$, then easy algebra shows
$$
x^2-4=\frac{(y+8)(y-4)}{9}
$$
and the given limit becomes
$$
\lim_{y\to4}\frac{f(y)-5}{y-4}\frac{1}{y+8}=1
$$
Since the limit is finite, we need $f(4)=5$; since $f'(4)=b$, the limit says
$$
\frac{b}{4+8}=1
$$
so $b=12$.
