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?