If $\lim \limits_{x \to 1} \frac{f(x)+2}{x-1} = 3$, compute: $\lim \limits_{x \to -1} \frac{(f(-x))^2-4}{x^2-1}$ Be $f(x)$ a polynomial in $\Bbb R$ such that:
If$$\lim \limits_{x \to 1} \frac{f(x)+2}{x-1} = 3$$
Compute: $$\lim \limits_{x \to -1} \frac{(f(-x))^2-4}{x^2-1}$$
I noticed that, if the first limit exists, then $f(x)+2 = (x-1)P(x)$, where $P(x)$ is another polynomial.
Then $$\lim \limits_{x \to 1} P(x) = 3$$
I tried to use that in the second limit, but i can't proceed further.
Any hints?

Given that $$\lim \limits_{x \to 1} \frac{f(x)+2}{x-1} = 3\tag1$$
Take $~f(x)=3x-5~$, then the above limit is satisfied.
Now $$\lim \limits_{x \to -1} \frac{(f(-x))^2-4}{x^2-1}=\lim \limits_{x \to -1} \frac{(-3x-5)^2-4}{x^2-1}$$
$$=\lim \limits_{x \to -1} \frac{(3x+3)(3x+7)}{x^2-1}$$
$$=\lim \limits_{x \to -1} \frac{3(3x+7)}{x-1}$$
$$=\frac{3(-3 +7)}{-1-1}=-6$$
