Proving this differentiable function $f: \mathbb{R}^+ \to \mathbb{R}$ is uniformly continuous Let $f: \mathbb{R}^+ \mapsto \mathbb{R}$ be a differentiable function such that $\lim_{x \to \infty} f'(x) = 1.$ Prove that $f$ is uniformly continuous on $\mathbb{R}^+$.
Attempt: We need to prove that $$\forall \epsilon > 0, \exists \delta > 0, \forall x, y \in \mathbb{R}^+: |x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon. $$
I think I almost found the whole proof, except the last case is still bothering me. I'd appreciate any feedback and/or help.

Since $\lim_{x \to \infty} f'(x) = 1$, there exists a $N > 0$ such that for all $x \in \mathbb{R}^+$ with $x > N$ we have that $$| f'(x) - 1 | < 1 $$ or $0 < f'(x) < 2. $ Now consider the interval $[0, N]$. The restriction of $f$ to this interval is an uniformly continuous function, since $f$ is then a continuous function defined on a closed and bounded interval. Let $\epsilon > 0$ be arbitrary. Let $\delta = \epsilon/2$. Then $\forall x,y \in [0, N]$ with $|x-y| < \delta$ we have that $$|f(x) - f(y)| < \epsilon.$$ Now we need to show the same $\delta > 0$ 'works' for all $x, y \in \mathbb{R}^+$. So  let $x, y \in \mathbb{R}^+$ be arbitrary. 


*

*Case 1: if $x,y \in [0, N]$, we are done.

*Case 2: Suppose that $x,y > N$. Suppose that $|x - y | < \delta$, and assume, without loss of generality, that $y < x$. Then the theorem of Lagrange says there exists a $c \in (y,x)$ such that $$f'(c) = \frac{f(x) - f(y)}{x-y}. $$  Since $c > N$, we have $f'(c) < 2. $ Also $x \neq y$. Hence $$ | f(x) - f(y)| = |f'(c)| | x-y| < 2\delta = \epsilon. $$

*Case 3: Suppose now that $y \in [0, N]$ and that $x > N$ (or the symmetric case where $x \in [0,N]$ and $y > N)$. Suppose that $|x - y| = x - y < \delta$. Then we also have $x - N < x - y < \delta$ since $N > y$. The continuous function $f$ is bounded on $[0,N]$. Hence there exists an $M > 0$ such that $\forall y \in [0,N]$ we have $|f(y)| \leq M$. Now consider the interval $[N,x]$. We can again find a $d \in (N,x)$ such that $$f'(d) = \frac{f(x) - f(N)}{x-N}. $$ Then it follows now that $$|f(x) - f(y)| \leq |f(x)| + |f(y)| < | f'(d)| | x - N | + |f(N)| \\ < 2 | x - y | + M < 2\delta + M. $$ 


I don't know how to do this last case properly, to get the desired $|f(x) - f(y)| < \epsilon. $ Do I really have to use Lagrange again (like I did), or does someone see a better and nicer way to prove the last case?
 A: Ok there is minor issue here. Normally the notation $\mathbb{R}^{+}$ is used to denote set of positive real numbers i.e. $(0, \infty)$. Then the function $f(x) = x + (1/x)$ provides a counterexample to the claim.
From the arguments given by OP in his question it is clear that the intention is to deal with $[0, \infty)$ and not $(0, \infty)$ and the same will be done in the answer below. And for sake of clarity I mention what needs to be proved.
If $f: [0, \infty) \to \mathbb{R}$ is a differentiable function with $\lim_{x \to \infty}f'(x) = 1$ then $f$ is uniformly continuous on $[0, \infty)$.
The idea is simple and works more generally when $\lim_{x \to \infty}f'(x) = L$ exists and we don't need $L$ to be necessarily equal to $1$. Since $f'(x) \to L$ as $x \to \infty$ it follows that there is a number $N > 0$ such that $$|f'(x) - L| < 1$$ for all $x > N$. Thus $f'(x)$ is bounded on $[N, \infty)$ and let $|f'(x)| < K$ for all $x \in [N, \infty)$ and some number $K$. This is the part which is needed to prove the uniform continuity. The existence of limit of $f'(x)$ is just one way to ensure that $f'(x)$ is bounded for large $x$.
Let $\epsilon > 0$ be given. Now $f$ is continuous on $[0, N]$ and therefore it is uniformly continuous on $[0, N]$. Therefore there is $\delta_{1} > 0$ such that if $x, y \in [0, N]$ and $|x - y| < \delta_{1}$ then $$|f(x) - f(y)| < \frac{\epsilon}{2} < \epsilon$$ Next if $x, y \in [N, \infty)$ then $$|f(x) - f(y)| = |f'(\xi)||x - y| < K|x - y|$$ via Mean Value Theorem. Let $\delta_{2} = \epsilon/2K$ and if $x, y \in [N, \infty)$ with $|x - y| < \delta_{2}$ then $$|f(x) - f(y)| < \frac{\epsilon}{2} < \epsilon$$ OP is handling these two cases perfectly well in his post and the problem occurs in handling the case when one of the points $x \in [0, N]$ and another point $y \in [N, \infty)$. This is not difficult and is based on adding and subtracting $f(N)$ to the difference $f(x) - f(y)$.
Now let $x, y$ be such that $x \in [0, N]$ and $y \in [N, \infty)$. And let $$|x - y| < \delta = \min(\delta_{1}, \delta_{2})$$ Since $N \in [x, y]$ it follows that $$|x - N| \leq |x - y| < \delta_{1}, |N - y| \leq |x - y| < \delta_{2}$$ Then
\begin{align}
|f(x) - f(y)| &= |f(x) - f(N) + f(N) - f(y)|\notag\\
&\leq |f(x) - f(N)| + |f(N) - f(y)|\notag\\
&< \frac{\epsilon}{2} + \frac{\epsilon}{2}\notag\\
&= \epsilon\notag
\end{align}
Thus we have found a $\delta > 0$ such that if $x, y \in [0, \infty)$ with $|x - y| < \delta$ then $|f(x) - f(y)| < \epsilon$. Therefore $f$ is uniformly continuous on $[0, \infty)$.
A: Let $\epsilon>0$. There is $\delta_1$ such that for all $x,y \in [0,N]$, $|x-y|<\delta_1 \implies |f(x)-f(y)|<\frac{\epsilon}{2}$ and there is $\delta_2$ such that for all $x,y \in [N,+\infty[$, $|x-y|<\delta_2 \implies |f(x)-f(y)|<\frac{\epsilon}{2}$
Let $x,y$ such that $|x-y|<\min(\delta_1,\delta_2)$.
If $x,y$ are as in case 1 and 2 then $|f(x)-f(y)|< \frac{\epsilon}{2}<\epsilon$.
For the case 3: $x \in [0,N]$ and $y \in [N,+\infty[$:
$|f(x)-f(y)| < |f(x)-f(N)| + |f(N)-f(y)| < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$ (because $|x-N|<\delta_1$ and $|y-N|<\delta_2$).
Hence $f$ is uniformly continuous on $\mathbb{R}_+$.
A: Your paragraph

"Now consider the interval $[0,N]$ The restriction of $f$ to this
  interval is an uniformly continuous function, since $f$ is then a
  continuous function defined on a closed and bounded interval. \ldots "

starts off correct (up to this point). You then write:

Let $\epsilon > 0$ be arbitrary. Let $\delta = \epsilon/2$. Then $\forall x,y \in [0,N]$ with
  $|x−y|<\delta$ we have that $|f(x)−f(y)|<\epsilon$."

This last sentence is wrong. You need to re-examine the definition of uniform continuity. 
That is to say: the problem in your proof is far earlier than the location you identified. 
