Find $\lim_\limits{x\to -1}{f(x)}$ 
Let $f:\mathbb{R}\mapsto\mathbb{R}$ be a function such that:
  
  
*
  
*$$f(x)=f(1-x), \forall x \in\mathbb{R}$$
  
*$$\lim_\limits{x\to 2}{\frac{f(x)+4}{x-2}}=1$$
Find $\lim_\limits{x\to -1}{f(x)}$.

I have tried the following:
$$\lim_\limits{x\to 2}{\frac{f(x)+4}{x-2}}=1\Leftrightarrow \lim_\limits{h\to 0}{\frac{f(2+h)+4}{h}}=1\Leftrightarrow \lim_\limits{h\to 0}{\frac{f(-h-1)+4}{h}}=1$$
So, I may need to show that $f(-h-1)=f(h+1)$ and I am done. Any hint?
 A: You are working too hard.  For the second condition to hold, it must be true that $$\lim_{x\to 2} f(x)+4=0$$
But for $x$ close to $2$, we have $1-x$ close to $1-2=-1$.  Hence $$\lim_{x\to 2}f(x)+4=\lim_{x\to -1}f(1-x)+4=\lim_{x\to -1}f(x)+4=0$$
A: For the limit in 2. to be finite and non-zero we must have $$ \lim_{x\to2} f(x)= -4$$ and thus, using 1. , $$\lim_{x\to-1} f(x) = -4.$$
A: We have $$1=\lim_{x\rightarrow2}\frac{f\left(x\right)+4}{x-2}=\lim_{x\rightarrow2}\frac{f\left(1-x\right)+4}{x-2}=\lim_{x\rightarrow-1}\frac{f\left(x\right)+4}{x+1}
 $$ then $$f\left(x\right)\sim x-3
 $$ at $x\rightarrow -1$ and so $$\lim_{x\rightarrow-1}f\left(x\right)=-4.
 $$
A: We can proceed as follows
\begin{align}
L &= \lim_{x \to -1}f(x)\notag\\
&= \lim_{x \to -1}f(1 - x) \text{ (property of }f(x))\notag\\
&= \lim_{t \to 2}f(t) \text{ (putting }t = 1 - x, t \to 2\text{ as }x \to -1)\notag\\
&= \lim_{x \to 2}f(x) \text{ (replacing }t \text{ with }x)\notag\\
&= \lim_{x \to 2}\{(f(x) + 4) - 4\}\notag\\
&= \lim_{x \to 2}\left\{\frac{f(x) + 4}{x - 2}\cdot(x - 2) - 4\right\}\notag\\
&= \lim_{x \to 2}\frac{f(x) + 4}{x - 2}\cdot\lim_{x \to 2}(x - 2) - 4\notag\\
&= 1\cdot 0 - 4 = -4\notag
\end{align}
