minimum and maximum problem I have some difficulties to solve this problem

Let $f\colon\mathbb R\to\mathbb R$ a continuous function such that
  $$\lim_{x \to-\infty}f(x)=\lim_{x \to+\infty}f(x).$$
  Prove that $ f $ has a minimum or a maximum in  $ \mathbb R $ (in the sense that it could also have them both).

 A: Write $\lim_{|x| \rightarrow \infty} f = l$. Fix $\epsilon$. By definition, away from an interval $I = [-R, R]$, $f$ lies within $\epsilon$ of $l$, and on $I$ $f$ is bounded by compactness, we deduce $f$ is bounded everywhere, hence $\inf f(\mathbb{R}), \sup f(\mathbb{R})$ exist.
From here it's case checking: if the $\inf$ and $\sup$ agree, $f$ is constant, if they disagree, $l$ is at most one of them, and since they are distinct they are separated by some epsilon, we deduce that the other must be obtained on an interval of the form $[-R, R]$ and by compactness is actually achieved by the function. 
A: If $f$ is constant, then it has a maximum and minimum trivially. Suppose $f$ is not constant, so we have $f(a)-f(b)>\epsilon$ for some $\epsilon>0,a,b\in \mathbb{R}$. Since $\lim_{x \to-\infty}f(x)=\lim_{x \to+\infty}f(x)$ we have some $M$ such that $x\geq M\implies |f(x)-\lim\limits_{x\to-\infty} f(x)|<\epsilon/4$ and some $N$ such that $x\geq N\implies |f(-x)-\lim\limits_{x\to+\infty} f(x)|<\epsilon/4$, so for $|x|>\max\{M,N\}$ we have $|f(-x)-f(x)|<\epsilon/2$. If we choose $x$ such that $|x|\geq a,b$ then we have that $f$ must attain a minimum or maximum on $[-|x|,|x|]$, as we have some point in between which is either above or below both endpoints.
A: Suppose that the common limit is the real number $c$.  Move $f$ up or down so that the limit is $0$. Call the resulting function $g$.
If $g$ is $0$ everywhere, we are finished. Otherwise, $g$ is either somewhere positive, or somewhere negative, or both.   
Suppose first that $g$ is positive somewhere. Then for some $a$, and some positive $\epsilon$, we have $g(a)=\epsilon$. 
There exists a real number $R$ such that $|g(x)|<\epsilon$ if $|x|>R$.  On the interval $[-R, R]$, the function $g$ attains a maximum value. That value is at least $\epsilon$, so it is greater than $g(x)$ for any $x$ such that $|x| >R$. It follows that $g$ attains a global maximum. 
If $g$ is somewhere negative, a mild modification of the above argument shows that $g$ attains a global minimum. Alternately, we can consider the function $-g$.
The result also holds if the limits are the same in the extended sense, for example if both limits are $+\infty$.  Pick any real $a$, and let $f(a)=b$. There is an $R$ such that $f(x)>b$ if $|x|>R$. On the interval $[-R, R]$, the function $f$ attains a minimum value.   That minimum value is $\le b$. Since $f(x)>b$ for $|x|>R$, that minimum value is a global minimum value. 
