Vtyjkyuk

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
$$ lim_xto 2 = f(3x-2)-5 over (x^2-4) = 9$$
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!

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
$$ lim_xto 2 fracf(3x-2)-5x^2-4 = lim_xto 2 frac3f'(3x-2)2x = frac34f'(4). $$

EDIT: If you don't want to use L'Hospital's rule, then you can also notice that
$$ lim_xto 2 fracf(3x-2)-5x^2-4 = lim_xto 2 frac1x+2 cdotlim_xto 2 fracf(3x-2)-5x-2 = frac14 lim_xto 2 fracf(3x-2)-f(4)x-2. $$
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
$$ g'(2) = 3f'(4). $$

Let $t = 3x-2$ so $x = fract+23$.

Then

$$9 = lim_xto 2fracf(3x-2)-5x^2-4 = lim_xto 2fracf(3x-2)-f(4)(x-2)(x+2) = frac14 lim_xto 2fracf(3x-2)-f(4)x-2 = frac14 lim_t to 4fracf(t)-f(4)fract-43 = frac34 f'(4)$$

If $3x-2=y$, then easy algebra shows
$$
x^2-4=frac(y+8)(y-4)9
$$
and the given limit becomes
$$
lim_yto4fracf(y)-5y-4frac1y+8=1
$$
Since the limit is finite, we need $f(4)=5$; since $f'(4)=b$, the limit says
$$
fracb4+8=1
$$
so $b=12$.