Proving that $\lim \limits_{x \to a^+} f(x)$ exists We have $f : (a,b) \to \mathbb{R}$ with the following property
$$|f(x) - f(y)| \leq M |x - y|^{1/2}$$
for $x, y \in (a,b)$ and a constant $M$. Prove that $\lim \limits_{x \to a^+} f(x)$ exists.

$f$ is uniformly continuous: fix $\epsilon > 0$, and take $\delta = \frac{\epsilon^2}{M^2}$.
Now, for any $\epsilon > 0$ there is a $\delta$ such that the $f(x)$ with $a < x < \delta$ differ by less than $\epsilon$. Thus it much approach some limit.
I believe my logic is correct, but I don't know how to make the argument more rigorous.
 A: You can't claim $f$ is uniformly continuous, given $f$ is continuous on an open interval $(a,b)$ (instead you would need the closed interval $[a,b]$). However, consider this. Let $x_n$ be a sequence in $(a,b)$ which converges to $a$. Then $|f(x_n) - f(x_m)| \le M|x_n - x_m|^{1/2} \to 0$ as $n,m\to \infty$. Hence $f(x_n)$ is a Cauchy sequence in $\Bbb R$. By completeness of $\Bbb R$, $\lim_{n\to \infty} f(x_n)$ exists. Let $L = \lim_{n\to \infty} f(x_n)$. Let $y_n$ be another sequence in $(a,b)$ which converges to $a$. Then
$$|f(y_n) - L| \le |f(y_n) - f(x_n)| + |f(x_n) - L| \le M|x_n - y_n|^{1/2} + |f(x_n) - L| \to 0$$
since $|x_n - y_n| \to 0$ and $|f(x_n) - L| \to 0$. This proves $f(a_n) \to L$ for every sequence $(a_n)$ in $(a,b)$ which converges to $a$. Therefore, $\lim_{x\to a^{+}} f(x) = L$.
A: Just about every type of limit has an associate Cauchy criterion.
In the present example, it states that $\displaystyle \lim_{x \to a^+} f(x)$ exists if, and only if, for every $\epsilon > 0$ there exists $\delta$ with the property that $$x,y \in (a, a+\delta) \implies |f(x) - f(y)| <\epsilon.$$
