Why $f(x)=\frac{1}{2}(x+\sqrt{x^2+1})$ is a pointwise contractive function? Let $(X,d)$ be a metric space and $f:X\to X$ a mapping on $X$.

$\textbf{Definition:}$ The mapping $f$ is said to be pointwise contractive if $\forall x\in X,$ there exists $\lambda_x\in [0,1)$ and an open neighborhood $U_x\subseteq X$ of $x$ s.t. $$d(f(x),f(y))\leq \lambda_x d(x,y)$$ for all $y\in U_x$.

How to prove that $f:\mathbb{R}\to \mathbb{R}, f(x)=\frac{1}{2}(x+\sqrt{x^2+1})$ is pointwise contractive mapping on the euclidean space?
 A: If $f$ is a $C^1$ function and $|f'(x)| < 1$ for every $x\in\mathbb{R}$, then a simple application of the mean value theorem shows that $f$ is a pointwise contraction.
Namely, let $x\in\mathbb{R}$, let $\delta>0$ and let $U_x := (x-\delta, x+\delta)$.
Since $f\in C^1$ and $|f'| < 1$, we have that
$$
\lambda_x := \max_{y\in [x-\delta, x+\delta]} |f'(y)| < 1
$$
and, by the mean value theorem,
$$
|f(x) - f(y)| \leq \lambda_x |x-y|
\qquad \forall y\in U_x.
$$
A: 
Yes according to your  definition $f$ is a point-wise contraction. Take $\color{blue}{ U_x=[x-1,x+1]}$ and $\color{red}{\lambda_x =\max\limits_{t\in[x-1, x+1]} f'(x) <1}$

Indeed,We have,
  $$f'(x)= \frac{1}{2} (1+\frac{x}{\sqrt{1+x^2}}) ~~\text{and}~~ f''(x) =\frac12(1+x^2)^{-3/2}>0 $$
  Hence, $x\mapsto f'(x )$ is a strictly increasing therefore, $$ 0\le f'(x) = \frac{1}{2} (1+\frac{x}{\sqrt{1+x^2}}) < \lim_{x\to \infty}\frac{1}{2} (1+\frac{x}{\sqrt{1+x^2}}) = 1$$
  That is for all,  $x\in\Bbb R$ we have, 
  $$\color{blue}{0\le f'(x) < 1}$$
Answer to the Question, for $x\in\Bbb R$ fixed, we have,
  we set $$\color{red}{\lambda_x =\max_{t\in[x-1, x+1]} f'(x) <1}$$
  Let $\color{blue}{y\in U_x=[x-1,x+1]}$
  Then we have, 
  $$|f(x)-f(y)| =\left|\int_x^yf'(t)dt\right|\le \lambda_x|x-y|$$

But it cannot be a contraction but rather it is a Lipschitz

Assume that $f$ is a contraction then there exists a constant $k<1$ such  that for all $ x\in \Bbb R$

$$|f(x) -f(0)|\le k|x|$$ which is impossible since 
 $$\color{red}{1 = \lim_{x\to \infty}\frac12\left| \frac{x-1}{x} +\frac{\sqrt{x^2+1}}{x}\right|=\lim_{x\to \infty}  \left|\frac{f(x)-f(0)}{x}\right|\le k< 1~~~\color{blue}{\text{**contradiction**}}}$$
Rather, $f$ is Lipschitz since we have,  $$|f'(x)|= \left|\frac{1}{2} (1+\frac{x}{\sqrt{1+x^2}})\right| \le 1$$
Hence, $|f'(x)\le 1$
Therefore, 
$$|f(x)-f(y)| =\left|\int_x^yf'(t)dt\right|\le |x-y|$$
