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!
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 $$ g'(2) = 3f'(4). $$