How does the existence of a limit imply that a function is uniformly continuous I am working on a homework problem from Avner Friedman's Advanced Calculus (#1 page 68) which asks

Suppose that $f(x)$ is a continuous function on the interval $[0,\infty)$. Prove that if $\lim_{x\to\infty} f(x)$ exists (as a real number), then $f(x)$ is uniformly continuous on this interval.

Intuitively, this argument makes sense to me. Since the limit of $f(x)$ exists on $[0,\infty)$, we will be able to find a $\delta > |x_0 - x_1|$ and this implies that, for any $\epsilon>0$, we have $\epsilon > |f(x_0) - f(x_1)|$ (independent of the points chosen). I am aware that the condition of uniform continuity requires that $\delta$ can only be a function of $\epsilon$, not $x$. 
What information does the existence of a real-valued limit provide that implies $f(x)$ is uniformly continuous on this interval?
 A: Remember the definition of "uniformly continuous":

$f(x)$ is uniformly continuous on $[0,\infty)$ if and only if for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that for all $x,y\in [0,\infty)$, if $|x-y|\lt \delta$, then $|f(x)-f(y)|\lt \epsilon$.

We also know that the limit exists. Call
$$\lim_{x\to\infty}f(x) = L.$$
That means that:

For every $\varepsilon\gt 0$ there exists $N\gt 0$ (which depends on $\varepsilon$) such that if $x\gt N$, then $|f(x)-L|\lt \varepsilon$.

Finally, you probably know that if $f(x)$ is continuous on a finite closed interval, then it is uniformly continuous on that interval.
So: let $\epsilon\gt 0$. We need to show that there exists $\delta\gt0$ such that for all $x,y\in [0,\infty)$, if $|x-y|\lt \delta$, then $|f(x)-f(y)|\lt\epsilon$.
We first use a common trick: if you know that any value of $f(x)$ in some interval is within $k$ of $L$, then you know that any two values of $f(x)$ in that interval are within $2k$ of each other: because if $|f(x)-L|\lt k$ and $|f(y)-L|\lt k$, then
$$|f(x)-f(y)| = |f(x)-L + L-f(y)| \leq |f(x)-L| + |L-f(y)| \lt k+k = 2k.$$
So: pick $N\gt 0$ such that for all $x\gt N$, $|f(x)-L|\lt \epsilon/2$. That means that if $x,y\gt N$, then $|f(x)-f(y)|\lt \epsilon$, by the argument above. So we are "fine" if both $x$ and $y$ are greater than $N$.
Now, we just need to worry about what happens if both $x$ and $y$ are in $[0,N]$, or if one of $x$ and $y$ is in $[0,N]$ and the other one is in $(N,\infty)$.
For both in $[0,N]$, we are in luck: since $f$ is continuous on $[0,\infty)$, then it is continuous on the finite closed interval $[0,N]$, hence is uniformly continuous there. So we know there exists $\delta_1\gt 0$ such that for all $x,y\in [0,N]$, if $|x-y|\lt\delta_1$, then we have $|f(x)-f(y)|\lt \epsilon$. So we just need to ensure that $x$ and $y$ are within $\delta_1$ of each other; that will ensure the inequality we want if $x$ and $y$ are both in $[0,N]$, or if they are both in $(N,\infty)$.
Now we run into a slight problem: what if, say, $x\in [0,N]$ and $y\in (N,\infty)$? Well, since $f$ is continuous at $N$, we know that we can ensure that $f(x)$ and $f(y)$ are both as close as we want to $f(N)$ provided that $x$ and $y$ are both very close to $N$. But if $x$ and $y$ are within some $\ell$ of $N$, then they are within $2\ell$ of each other (same argument as before); and if $f(x)$ and $f(y)$ are both within some $k$ of $f(N)$, then they will be within $2k$ of each other.
So: let $\delta_2$ be such that if $|a-N|\lt\delta_2$, then $|f(a)-f(N)|\lt \epsilon/2$. Then, if $x$ and $y$ are both within $\delta_2$ of $N$, then $|f(x)-f(y)|\lt \epsilon$, and we'll be fine.
In summary: we want to select a $\delta\gt 0$ that will ensure that if $|x-y|\lt\delta$, then:


*

*If $x$ and $y$ are both less than $N$, then $|x-y|\lt \delta_1$;

*If $x$ and $y$ are both greater than $N$, then it doesn't matter how close to one another they are; and

*If one of $x$ and $y$ is less than $N$ and the other is larger than $N$, then they are each within $\delta_2$ of $N$.


To make sure the first condition happens, we just need to make sure that $\delta\leq\delta_1$. The second condition is easy. What should we require of $\delta$ in order for the second condition to hold? If we can find a $\delta$ that makes all three things happens simultaneously, we'll be done.
A: If you've learned that continuous functions on compact sets are uniformly continuous, then this turns out to be a simple exercise with the extended real numbers.
We can extend $f$ by continuity to a function $f^*$ defined on the interval $[0, +\infty]$. That is, define
$$ f^*(x) = \begin{cases} f(x) & x < +\infty
\\ \lim_{y \to +\infty} f(y) & x = +\infty \end{cases} $$
Since $[0, +\infty]$ is compact, and $f^*$ is continuous, we can conclude $f^*$ is uniformly continuous on $[0, +\infty]$. And thus $f^*$ is uniformly continuous on $[0, +\infty)$ as well, from which we conclude $f$ is uniformly continuous.
A: This result is deeper and  has a simpler proof in general Euclidean spaces or general metric spaces. In fact let's  the result for Euclidean spaces.

Lemma If $f:\mathbb R^d\to \mathbb R$ is a continuous function such that $\lim_{|x|\to\infty } f(x) = L$ with $L\in\mathbb R$, then $f$ is uniformly continuous on $\mathbb R^d.$ 

Proof of the Lemma
Let $\varepsilon >0$. We want to show there is $\delta>0$ such that $ |f(x)-f(y)|\le \varepsilon$ whenever $|x-y|\le \delta$.
But since $\lim_{|x|\to\infty } f(x) = L$  we know that there is $R>0$ such that 
$$
\begin{equation}\label{eq1}\tag{I}|f(x)-L|\le \varepsilon/2 \qquad\text{whenever}\qquad |x|\ge R\end{equation}
$$
But since $f$ is continuous it is uniformly continuous on the compact $$\bar{B}(0,R+1) = \{x\in\mathbb R^d: |x|\le R+1\}$$

Hein Borel result  Every function $f$ continuous on compact set is therein uniformly continuous.

hence there exists $0<\delta<1$ such that for every 
\begin{equation}\label{eq2}\tag{II}
\text{$x,y\in \bar{B}(0,R+1)$ with $|x-y|<\delta$ we have $ |f(x)-f(y)|\le \varepsilon /2$}
.
\end{equation}

Now let $x,y\in \mathbb R^d$ such that $|x-y|<\delta$
  
  
*
  
*If $x,y\notin \bar{B}(0,R)$ i.e $|x|>R$ and $|y|>R$ by $\eqref{eq1}$ it readily follows that 
  $$|f(x)-f(y)|\le|f(x)-L|+|f(y)-L|\le \varepsilon .$$
  
*If $x,y\in \bar{B}(0,R)\subset\bar{B}(0,R+1)$ then \eqref{eq2} yields
  $$|f(x)-f(y)|\le \varepsilon .$$
  
*If $x \in \bar{B}(0,R)$ and $y\notin \bar{B}(0,R)$ then $y\in \bar{B}(0,R+1)$ indeed
  $$|y|\le|x|+|x-y|\le R+\delta <R+1$$
  thus \eqref{eq2} yields 
  $$|f(x)-f(y)|\le \varepsilon .$$
  

In any case we have  $ |f(x)-f(y)|\le \varepsilon$for all $x,y\in\mathbb R^d$ such that $|x-y|\le \delta$. Which prove the uniform continuity.

The case where $ \mathbb R^d$ is replaced by a locally compact metric space $(X,d)$ analogious reason lead to same conclusion. except that, $ f:(X,d)\to \mathbb R$ vanish at infinty means: for a given $ \varepsilon > 0$ there exists a Compact subset $K$ of $X$ such that $|f(x)|\le \varepsilon $ for every $x\in X\setminus K.$ Here the ball $\bar{B}(0,R+1)$ is replaced by the compact set . $$K_1 = \{x\in X: d(x,K)\le 1\}$$

A: We know that for all $\varepsilon > 0$ there exists $X \in \mathbf R$ such that for all $x \geqslant X$ we have $|f(x) - \ell| < \varepsilon$ where $\displaystyle \ell = \lim_{x \to \infty} f(x)$.
So pick $\epsilon > 0$. Then we get from the previous condition a real number $X_\varepsilon > 0$. $f$ is uniformly continuous on $[0, X_\varepsilon]$ because that interval is compact.
Now, on $(X_\varepsilon, \infty)$ we have $|f(x) - \ell| < \varepsilon$. So we will always have $|f(x) - f(y)| \leq 2\varepsilon$ for $x, y$ in $(X_\varepsilon, \infty)$. Can you finish this?
A: Consider for example the function $\tan : [0,\pi/2]\to [0,\infty]$ with the convention $\tan(\pi/2)=\infty$. This function is increasing and $C^\infty$ on $(0,\pi/2)$. 
Then you may consider $g:[0,\pi/2] \to \Bbb{R}$ defined by
$$ g(x)= \begin{cases} f(\tan x), & x \in [0,\pi/2) \\
\lim_{x \to \infty}f(x)=f(\infty) & x=\pi/2\end{cases}$$
Then $g$ is continuous on a compact set, therefore it is uniformly continuous. We can obtain $f$ by using the composition $f(x)=g(\arctan x)$. We know that $(\arctan x)'=\frac{1}{1+x^2}\leq 1$, which means, by the intermediate value theorem, that $|\arctan x-\arctan y| \leq |x-y|$ for every $x,y \in [0,\infty)$. Now pick $\varepsilon >0$ in the uniform continuity of $g$. Then there exists $\delta >0$ such that $|x-y|<\delta \Rightarrow |g(x)-g(y)|<\varepsilon$. But then $|\arctan x-\arctan y|\leq |x-y|<\delta$, therefore 
$$ |f(x)-f(y)|=|g(\arctan x)-g(\arctan y)|<\varepsilon $$
This means that for every $\varepsilon >0$ there exists $\delta$ (the same as in the uniform continuity of $g$) such that every $x,y \in [0,\infty)$ with $|x-y|<\delta$ it follows that $|f(x)-f(y)|<\varepsilon$. Therefore $f$ is uniformly continuous.

What I did above was just translating the structure of the space $[0,\infty]$ which is compact, to a usual compact interval. The condition that $f$ has a limit at $\infty$ means that $f$ is continuous on the space $[0,\infty]$, which is the compactification of $[0,\infty)$ by adding another point, namely $\infty$. Why is $[0,\infty]$ compact? 


*

*if $(y_n) \subset [0,\infty]$ then either $(y_n)$ has a bounded subsequence which by the Weierstrass theorem implies that there is a convergent subsequence, either $(y_n)$ is unbounded, which means that there is a subsequence converging to $\infty$.
Then the theorem that says that any continuous function on a compact set is uniformly continuous can be applied. The arguments above are a workaround this.
