Finding $ \lim_{x\to1/2}\frac{f(x)}{x-1/2}$ given that $ \lim_{x\to1}\frac{f(x)}{x-1}=2$ and $\lim_{x\to-1}\frac{f(x)}{x+1}=6$ Consider the following question:

Find $\displaystyle \lim_{x\to1/2}\frac{f(x)}{x-1/2}$
given that $\displaystyle \lim_{x\to1}\frac{f(x)}{x-1}=2$ and
$\displaystyle \lim_{x\to-1}\frac{f(x)}{x+1}=6$

Attempt
Let $f(x)=(x-1)(x+1)(x-1/2)q(x)$. Here I have $q(1)=2$ and $q(-1)=2$.
\begin{align}
\lim_{x\to1/2}\frac{f(x)}{x-1/2} 
&=\lim_{x\to1/2}\frac{(x-1)(x+1)(x-1/2)q(x)}{x-1/2}\\
&=\lim_{x\to1/2}\left((x-1)(x+1)q(x)\right)\\
&= -\frac{3}{4}q(1/2).
\end{align}
Question
Is it possible to find $q(1/2)$?
 A: Conlusion: from what's known, we can't find the limit. We can make some weak assumptions which doesn't lose the generality. If $f \in C^2$, or f is continously differential, then we can use the L.Hospital Laws. 
We can have
$\quad$ $f'(1)=2$ and $f'(-1) =6$. But we get nothing about $f'(1/2)$.
A: Not enough information given. 
For example, the polynomial $f(x)=ax^4+(1-2a)x^2+a-1$ with $a \in \mathbb{R}$ checks the hypothesis. The limit though, is finite only when $f\left(\frac{1}{2}\right) = 0$, which gives $a = \frac{4}{3}$ and the limit is $-1$,otherwise the limit can be $\pm \infty$. If we go for a polynomial of higher degree, similar discussion arises. Not to mention there are more complex functions that can satisfy the requirements.
If we're given that $f$ is a third degree polynomial, then we can solve the question:
$$f(x)=ax^3+bx^2+cx+d$$
and we need:
$$f(1)=a+b+c+d=0$$
$$f(-1)=-a+b-c+d=0$$
$$f'(1)=3a+2b+c=2$$
$$f'(-1)=3a-2b+c=6$$
Solving the system gives $f(x)=2x^3-x^2-2x+1=(x-1)(x+1)(2x-1)$, and the limit is finite:
$$\lim_{x\to \frac{1}{2}}\frac{f(x)}{x-\frac{1}{2}}=\lim_{x\to \frac{1}{2}}2(x-1)(x+1)=-\frac{3}{2}$$
