Find $\lim_\limits{x\to -1}{f(x)}$

Question

Let $f:\mathbb{R}\mapsto\mathbb{R}$ be a function such that:

1. $$f(x)=f(1-x), \forall x \in\mathbb{R}$$

2. $$\lim_\limits{x\to 2}{\frac{f(x)+4}{x-2}}=1$$

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?

Answer 1

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 $$\lim_{x\to 2}f(x)+4=\lim_{x\to -1}f(1-x)+4=\lim_{x\to -1}f(x)+4=0$$

Answer 2

For the limit in 2. to be finite and non-zero we must have $$ \lim_{x\to2} f(x)= -4$$ and thus, using 1. , $$\lim_{x\to-1} f(x) = -4.$$

Answer 3

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 $$\lim_{x\rightarrow-1}f\left(x\right)=-4. $$

Answer 4

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}