Lipschitz function problem I can not conclude this problem, I tried two ways but I can not conclude ...
Let,
$$f:(0,+\infty)\rightarrow \mathbb{R}$$ Lipschitz function.
Prove that there exists a finite limit:
$$\lim_{x \to 0^+}f(x)$$
FIRST ATTEMPT:
Let $x_0$ and $x$ two points of the domain of $f$:
$$x_0 , x \in (0, + \infty).$$
We know the function is Lipschitz continuous (and thus also continuous): by definition Lipschitz functionwe have:
$$|f(x)-f(x_0)|\le L| x-x_0 | $$
from which:
$$ \frac {| f (x)-f (x_0)|}{| x-x_0 |} \le L $$
Then, passing to the limit $x \to x_0$
$$\lim_ {x \to x_0} \frac {| f (x)-f (x_0 )|}{| x-x_0 |} = f '(x) \le L $$
but from here I can not prove $\lim_{x \to 0^+}f(x)$
SECOND ATTEMPT:
$$ \forall x, y \in(0, + \infty), \quad L| x-y | \le f(x)-f (y) \le L | x-y |$$
or:
$$ L | x-y | + f(y) \le f(x) \le L | x-y | + f (y) $$
$$ f (x) \le L | x-y | + f (y) $$
passing to the limit $ x \to 0 ^+ $
$$\lim_ {x \to x_0} f (x) \le L | x-y | + f (y) $$
but from here can I  conclude that $\lim_{x \to 0^+}f(x)$ is a finite limit?
 A: I'll try to say something about your attempts in a bit, but here's another idea.
Let $\{x_n\}$ be a sequence in $(0, \infty)$ such that $x_n \to 0$. You can use the Lipschitz condition to show that the sequence $\{f(x_n)\}$ is Cauchy and hence convergent to a limit $p$. If we select another sequence $\{y_n\}$ in $(0, \infty)$ such that $y_n \to 0$ and $f(y_n) \to q$, then we'd like to show that $p = q$.
For this, assume that $\varepsilon = |p - q|$ is non-zero and choose an integer $N$ such that $n \geq N$ implies the following:


*

*$|f(x_n) - p| < \varepsilon/3$ and $|f(y_n) - q| < \varepsilon/3$

*$x_n, y_n < \varepsilon/(3K)$, where $K$ is a Lipschitz constant for $f$.

A: Another way to look at the liminf/limsup approach is to do it as a proof by contradiction... if the limit did not exist, then $\limsup_{x \rightarrow 0+}f(x) = B$ and $\liminf_{x \rightarrow 0+}f(x) = A$, where $A< B$. (Lipschitz functions on $(0,1)$ are immediately bounded so you don't have to worry about $A$ or $B$ being $\pm \infty$.) Then for some small $\epsilon > 0$ you could find $a_n \rightarrow 0$ and $b_n \rightarrow 0$ such that $f(a_n) < A + \epsilon$ and $f(b_n) > B - \epsilon$. So $|f(b_n) - f(a_n)| > B - A - 2\epsilon$, even though $|b_n - a_n|$ goes to zero, which contradicts Lipschitzness. 
And you can do this less rigorously without any liminfs or limsups. If the limit doesn't exist, then there are $a_n,b_n \rightarrow 0$ such that $|f(b_n) - f(a_n)| > \delta$ for some fixed $\delta$, despite $|b_n - a_n|$ going to zero. The liminf and limsup just formalizes this idea.
