Show that $f$ is uniformly continuous - $\lim_{\|x\| \to \infty}f(x) = c$ 
Let $f : \mathbb{R}^n \to \mathbb{R}^m$ be a continuous function. We
  suppose that there exist $c \in \mathbb{R}^m$ such that $$\lim_{\|x\| 
 \to \infty} f(x) = c.$$ Show that $f$ is uniformly continuous.

We know that if $\forall \epsilon_1 > 0$, $\exists M > 0$ such that $\forall x \in \mathbb{R}^n\setminus B(0,M) $, we have $\|f(x) - c\| < \epsilon_1$.
We want to show that $\forall \epsilon_2 > 0$, $\exists \delta > 0$ such that $\forall x,\bar{x} \in \mathbb{R^n} $, we have $\|f(x) - f(\bar{x})\| < \epsilon_2$ as long as $\|x-\bar{x}\|<\delta$.
Let $x,\bar{x} \in \mathbb{R^n} $ and $\|x-\bar{x}\|<\delta$.
Take $\epsilon_1 = \frac{\epsilon_2}{2} $
$\implies \|f(x) - c\| < \frac{\epsilon_2}{2}$ as long as $\|x\|\geq M$ and $\|f(\bar{x}) - c\| < \frac{\epsilon_2}{2}$ as long as $\|\bar{x}\|\geq M$
I am stuck on this point. Which $\delta$ could I take to prove the uniform continuity of $f$?
 A: The thing you have shown above proves the uniform continuity of $f$ outside the ball. Actually, it shows that there exist $M>0$ such that for any two points outside the ball of radius $M$, $$\|f(x)-f(y)\|<\epsilon_2.$$
Now, we need to show that continuous function $f$ is uniformly continuous for the ball $B(0,M)$. 
Since the closure of $B(0,M)$ is closed, using the Heine–Cantor theorem we can say that $f$ is uniformly continuous for the the ball $B(0,M)$. 
That proves your statement.
A: Let $\epsilon>0$ is arbitrary.
From $\lim\limits_{\|x\|\to\infty}{f(x)}\to c$ it follows that $\forall \epsilon>0,\,\exists M>0:\|x\|\ge M \Rightarrow \|f(x)-c\|\leq \epsilon$. Chose this $M$, corresponding to our $\epsilon$ devided by $2$. In the ball $\|x\|\leq M+1$ (this $+1$ is to take care of the points $x,y$ from each side of the boundary of $B(0,M)$)the function $f$ is uniformly continuous, as a continuous function over the compact set $B(0,M+1)=\{x\in\mathbb R^n:\|x\|\leq M+1\}$. It is left to show that $f$ is uniformly continuous in $\mathbb R^n\setminus B(0,M)$. Here we use that $\|f(x)-c\|\leq \frac{\epsilon}{2}$:
Let $x,y\in\mathbb R^n\setminus B(0,M)$. Then $\|f(x)-f(y)\|\leq \|f(x)-c\|+\|f(y)-c\|\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.
Finally, from the uniform continuity in $B(0,M+1)$ for the arbitrary $\epsilon>0$ we have $\exists\delta_1>0: x,y\in B(0,M+1):\|x-y\|\leq \delta_1\Rightarrow \|f(x)-f(y)\|\leq \epsilon$. The $\delta$ for the whole $\mathbb R^n$ we chose as $\delta:=\min{\{\delta_1,1\}}$.
