extend an alternative definition of limits to one-sided limits The following theorem is an alternative definition of limits. In this theorem, you don't need to know the value of $\lim \limits_{x \rightarrow c} {f(x)}$ in order to prove the limit exists.

Let $I \in R$ be an open interval, let $c \in I$, and let $f: I-{c} \rightarrow R$ be a function. Then $\lim \limits_{x \rightarrow c} {f(x)}$ exists if for each $\epsilon > 0$, there is some $\delta > 0$ such that $x,y \in I-{c}$ and $\vert x-c \vert < \delta$ and $\vert y-c \vert < \delta$ implies $\vert f(x)-f(y) \vert < \epsilon$.

I have extended this to a theorem for one-sided limits: $\lim \limits_{x \rightarrow c+} {f(x)}$ exists if or each $\epsilon > 0$, there is some $\delta > 0$ such that $c<x<c+ \delta$, and $c<y<c+ \delta$ implies $\vert f(x)-f(y) \vert < \epsilon$.
My question is: how to prove this extension to one-sided limits? I'm not familiar with Cauchy sequences, so I would prefer $\epsilon$-$\delta$ proof. 
 A: You have essentially the same difficulty as in the previous question: figuring out a "target" number $L$ for the limit.
(By the way: if you don't know what Cauchy sequences are, you should have said something in that previous problem!)
You can proceed in a similar manner. First, you want to find a potential "target." For $\epsilon_n = \frac{1}{n}$, you know there is a $\delta_n$ (which you may assume is less than or equal to $\frac{1}{n}$ and less than $\delta_{n-1}$) such that for all $x$ and $y$, if $c\lt x\lt c+\delta_n$ and $c\lt y \lt c+\delta_n$, then $|f(x)-f(y)|\lt \frac{1}{n}$. 
Note that this means that the values of $f$ on $(c,c+\delta_n)$ are bounded: for any $x\in (c,c+\delta_n)$, you know that $|f(x) - f(c+\delta_n/2)|\lt \frac{1}{n}$, so $|f(x)| \lt \frac{1}{n}+|f(c+\delta_n/2)|$. 
In particular, there is a greatest lower bound $a_n$ to $f(c,c+\delta_n)$. Since $(c,c+\delta_{n+1})\subseteq (c,c+\delta_n)$, we have that $a_n\leq a_{n+1}$. So the sequence $a_1,a_2,\ldots$ is an increasing function. The sequence is bounded above, so the sequence converges to some $L$.
The obvious "target" for the limit is $L$. So, let $\epsilon\gt 0$, and you want to show that there is a $\delta\gt 0$ such that for all $x$, if $c\lt x\lt c+\delta$, then $|f(x)-L|\lt \epsilon$.
Since $L$ is the least upper bound of $a_1,a_2,\ldots$, there exists $N$ such that $L-\frac{\epsilon}{2} \lt a_n \leq L$ for all $n\geq N$. And there exists an $M$ such that $\frac{1}{M}\lt \frac{\epsilon}{4}$. Let $K=\max\{M,N\}$, and consider $\delta=\delta_K$.
We know that since $a_K$ is the greatest lower bound of $f(c,c+\delta_K)$, there exists $y\in (c,c+\delta_K)$ such that $a_K\leq f(y)\lt a_K+\frac{\epsilon}{4}$. 
Now suppose that $c\lt x \lt c+\delta_K$. Then
\begin{align*}
|f(x)-L| &\leq |f(x)-f(y)|+|f(y)-a_K| + |a_K-L|  &&\mbox{(triangle inequality)}\\\
&\lt \frac{1}{K} + |f(y)-a_K| + |a_K-L| &&\mbox{(choice of $\delta_K$)}\\\
&\leq \frac{1}{K} + \frac{\epsilon}{4} + |a_K-L| &&\mbox{(choice of $y$)}\\\
&\leq \frac{1}{K} + \frac{\epsilon}{4} + \frac{\epsilon}{2} &&\mbox{(since $K\geq N$)}\\\
&\lt \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{2} &&\mbox{(since $K\geq M$)}\\\
&=\epsilon.
\end{align*}
(With this in mind, you can try doing the two-sided version of the problem without invoking Cauchy sequences; the idea is very similar, and you don't need to consider both least upper bounds and greatest lower bounds for $f(c-\delta_n,c+\delta_n)$, just one of the two and use their limit as a "target". Once you know what to aim for, hitting it becomes much easier).
A: Answer: For each positive $\epsilon$, there is some positive $\delta$ such that  $\vert f(x)-f(y) \vert \le \epsilon$ for every $x$ and $y$ in $I$ such that $c<x\le c+\delta$ and $c<y\le c+\delta$.
About the question added later on in your post: you could try to mimick the proof given in the two sided case.
A: Assume that $\lim_{x \to c^+} f(x) = g$ exists. Then for any $\varepsilon > 0$ there exists $\delta > 0$ such that if $0 < x - c < \delta$ then $|f(x) - g| < \varepsilon$ and if $0 < y - c < \delta$ then $|f(y) - g| < \varepsilon$. Hence
$$
|f(x) - f(y)| \leq |f(x) - g| + |f(y) - g| < 2\varepsilon
$$
for $0 < x - c < \delta$ and $0 < y - c < \delta$.
On the other hand, assume that your definition holds. Then there exists a strictly decreasing sequence $(c_n)_{n \geq 1}$ such that $c_n \to c$. Thus for a given $\varepsilon > 0$ there exists $\delta > 0$ such that there exists positive integer $N$ such that $|f(c_m)-f(c_n)| < \varepsilon$ for $m,n > N$.
It means that $(f(c_n))$ is Cauchy sequence and converges to, say, $g$.
(If you are not familiar with Cauchy sequences, it is easy to prove that the sequence $(a_n)$ such that for $\varepsilon > 0$ there exists $N > 0$ such that $|a_m - a_n| < \varepsilon$ for $m,n > N$ converges. To prove it, notice first that $(a_n)$ is bounded. Then, by Bolzano-Weierstrass theorem there is a subsequence $(a_{n_k})$ which converges to some $a$. Then, by Cauchy condition and triangle inequality $a_n \to a$.)
Assume now that there exists a sequence $(c'_n)_{n \geq 1}$ such that $f(c'_n) \to g' \neq g$. Let $C_n$ equals $c_n$ if $n$ is even, and equals $c'_n$ if $n$ is odd. Hence $(f(C_n))$ diverges but on the other hand we have $|f(C_m) - f(C_n)| < \varepsilon$ for sufficiently large $m,n$ and this means that $(f(C_n))$ is Cauchy, hence converges. Contradiction.
