Cauchy condition for functions 
Prove that $f$ has a limit at $a$ if and only if for every $\epsilon > 0$, there exists $\delta>0$ such that if $0<|x-a|<\delta$ and $0<|y-a|<\delta$, then $|f(x)-f(y)|<\epsilon$.

Forward direction: Suppose $f$ has  a limit $L$ at $a$. Fix $\epsilon$. Then for some $\delta$ we have $|f(x)-L|<\epsilon/2$ whenever $|x-a|<\delta$. Then for $|x-a|,|y-a|<\delta$, we have $|f(x)-f(y)|\le|f(x)-L|+|f(y)-L|<\epsilon/2+\epsilon/2=\epsilon$.
Backward direction: Suppose there exists $\delta>0$ such that if $0<|x-a|<\delta$ and $0<|y-a|<\delta$, then $|f(x)-f(y)|<\epsilon$. We want to show $f$ has limit at $a$, which means that for some $L$, any sequence of $x_i$'s converging to $a$ has $f(x_i)$'s converging to $L$. How to proceed?
 A: Suppose that $(x_n)$ converges to $a$. By the Bolzano-Weierstrass theorem, we know that the sequence $(f(x_n))$ must have a monotone subsequence $(f(x_{n_j}))$. 
Further, $(f(x_{n_j}))$ must be bounded: taking $\epsilon=1$, there exists $\delta>0$ so that $\lvert x-a\rvert<\delta$ and $\lvert y-a\rvert<\delta$ implies $\lvert f(x)-f(y)\rvert<1$; in particular, for all $x\in(a-\delta,a+\delta)$ we have $\lvert f(x)-f(a+\frac{\delta}{2})\rvert<1$, which implies
$$
\lvert f(x)\rvert<\lvert f(a+\tfrac{\delta}{2})\rvert+1\text{ for all }x\in(a-\delta,a+\delta);
$$
since $x_{n_j}$ is eventually contained in $(a-\delta,a+\delta)$, the sequence $(f(x_{n_j}))$ is then bounded.
So, $(f(x_{n_j}))$ is bounded and monotone, and therefore converges to some $L$. We claim that $(f(x_n))$ must converge to $L$ as well.
Let $\epsilon>0$ be given.  By assumption, there exists $\delta>0$ such that $0<\lvert x-a\rvert<\delta$ and $0<\lvert y-a\rvert<\delta$ implies $\lvert f(x)-f(y)\rvert<\frac{\epsilon}{2}$. Because $x_n\rightarrow a$, there exists $N\in\mathbb{N}$ so that $n>N$ implies $\lvert x_n-a\rvert<\delta$. Because $x_{n_j}\rightarrow a$ and $f(x_{n_j})\rightarrow L$, there exists $J\in\mathbb{N}$ such that $\lvert x_{n_J}-a\rvert<\delta$ and $\lvert f(x_{n_J})-L\rvert<\frac{\epsilon}{2}$.
Then for $n>N$, 
$$
\lvert f(x_n)-L\rvert\leq\lvert f(x_n)-f(x_{n_J})\rvert+\lvert f(x_{n_J})-L\rvert<\epsilon.
$$
So, $f(x_n)\rightarrow L$, as claimed.
A: Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces and $f:X \setminus a  \to Y$  ($a$ could be in the domain of $f$, but doesn't need to be).
Let $a \in X$ be a limit point of X. Then there is at least one sequence $(a_i)$ of points  of X distinct from $a$ converging to $a$, and the sequence is in the domain of $f$.
Let $Y$ be complete so that all Cauchy sequences have a limit point in Y.
Note that for a real function $f: \mathbb R \to  \mathbb R$ with the modulus metric $|.|$ these conditions are satisfied, as all points are limit points and $\mathbb R $ is complete.
$f$ having a limit $L$ at $a$ is defined as the function $g$ being continuous at $a$ where $g(a) = L$ and $g(x \ne a) = f(x)$.
In a metric space continuity is equivalent to sequential continuity, i.e. "$g$ is continuous at $a$" is equivalent to "for all sequences $(a_i)$ (of which we have at least one) which converge to $a$ then $g(a_i)$ converges to $g(a)$".
Proof:
(1) We take that given $\epsilon$ there exists $\delta$ such that for $d_X(a, x_1),  d_X(a, x_2) < \delta$ then $d_Y(f(x_1), f(x_2)) < \epsilon$.  
Let $A = (a_i)$ be a sequence in $X$ converging to $a$.
Then there is $N_A$ such that for $i > N_A$ we have $d_X(a, a_i) < \delta$.
Then let $i, j > N_A$, so by (1),  $d_Y(f(a_i), f(a_j)) < \epsilon$ which is the condition that $(f(a_i))$ is a Cauchy sequence in $Y$.
Since $Y$ is complete then the sequence $(f(a_i))$ has a limit in $Y$ say $L$ and we define $g(a) = L$.  
It remains to show that for any other sequence $(b_i)$ which converges to $a$ then $(f(b_i))$ converges to the same limit $L$.
Let $N_B$ be such that for $k > N_B$ we have $d_X(a, b_k) < \delta$
Since $(f(a_i))$ converges to $ L$ we can find $N$ such that for $i > N$ we have $d_Y(L, f(a_i)) < \epsilon$
Then let $M = $max$(N, N_A, N_B)$ and for $i, k > N$ we have $d_Y(f(a_i), f(b_k)) < \epsilon$  and $d_Y(L, f(a_i)) < \epsilon$
So by triangle inequality, $d_Y(L, f(b_i)) \le d_Y(L, f(a_i)) + d_Y(f(a_i), f(b_k)) = 2\epsilon$
I.e. $(f(b_i))$ converges to the same limit $L$.  
So the function $g$ is sequentially continuous at $a \in X$ and therefore continuous which defines $L$ as the limit of $f$ at $a$.
A: Let me propose an alternate proof for this direction.
We shall use Heine definition of Limit:
We shall say $\lim_{x\to a} f(x) = L$ if and only if for every $(x_n)_{n=1}^\infty $ such that for all n $x_n \ne a$ and $x_n \xrightarrow[n \to \infty]{} a$  we have $f(x_n) \xrightarrow[n \to \infty]{} L$
So let there be such $(x_n)_{n=1}^\infty$ and $\epsilon > 0$, from our assumption that the Cauchy criterion holds, there exist $\delta > 0$ such that for all $x,y \in (a-\delta,a+\delta)$ we have $|f(x)-f(y)| < \epsilon$.
so from our assumption that $x_n$ converges to $a$ there exist $n_0\in \Bbb{N}$ such that for all $n > n_0$, $x_n \in (a-\delta,a+\delta)$. Now, for all $m,n > n_0$ we have $x_m,x_n \in (a-\delta,a+\delta)$, so we also have $|f(x_m)-f(x_n)| < \epsilon$  so $f(x_n)$ is a Cauchy sequence and thus is convergent.
So we proved that for every $(x_n)$ that converges to $a$, $f(x_n)$ converges. now we have to prove that all such sequences converge to the same limit:
Let there be $(y_n)$ and $(x_n)$ that converge to $a$, then $f(x_n)$ and $f(y_n)$ converge to some $L,K \in \Bbb{R}$ respectively.
let us define $(z_n)$ so it will have $x_n$ in its even indices and $y_n$ in its odd indices. It is clear that $\lim_{n \to \infty}z_n = a$ so that means that $f(z_n)$ converges, let's call its limit $P \in \Bbb{R}$.
So because clearly $f(x_n)$ and $f(y_n)$ are subsequences of $f(z_n)$, then they also converge to $P$. by the uniqueness of the limit that means $L=K$.
So thus we proved Heine definition of the limit, and we proved that if the Cauchy criterion for function is met (at $a \in \Bbb{R}$), then the function has a limit at $a$.
A: Cauchy Convergence Condition for Functions
Prove that f has a limit at a if and only if for every ϵ > 0, there exists δ > 0  such that if 0 < |x−a| < δ  and 0 < |y−a| < δ , then |f(x)−f(y)| < ϵ.
Proof
Forward direction: Suppose f has a limit L at a. Fix ϵ. Then for some δ we have |f(x)−L| < ϵ/2 whenever |x−a| < δ. Then for |x−a|, |y−a| < δ, we have |f(x)−f(y)| ≤  |f(x)−L| + |f(y)−L| < ϵ/2 + ϵ/2 = ϵ.
Backward direction: Suppose there exists δ > 0 such that if 0 < |x−a| < δ and 0 < |y−a| < δ, then |f(x)−f(y)| < ϵ. We want to show f has limit at a, which means that for some L, any sequence of xi's converging to a has f(xi)'s converging to L.
Suppose that (xn) converges to a.
Let ϵ > 0 be given. By assumption, there exists δ > 0 such that 0 < |x−a| < δ and 0 < |y−a| < δ implies |f(x) − f(y)| < ϵ. Because xn→ a, there exists N so that n > N, m > N implies |xn−a| < δ and |xm−a| < δ which implies |f(xn) − f(xm)| < ϵ.
But by Cauchy Convergence Condition for Sequences:
{an} is convergent iff ϵ > 0 there exists N ∈ N so that n > N, m > N implies |an − am| < ϵ.
Thus {f(xn)} converges. We now prove that all such sequences converge to the same limit point as required by theorem:
lim x -> a f(x) = L iff  lim n -> ∞ f(xn) = L for all sequences {xn} such that lim n -> ∞ xn = a.
Consider any two such sequences{xn} and {yn} where yn→ a and {f(yn)} converges.  Then for a given ϵ > 0 there exists N so that n > N, m > N implies |xn−a| < δ and |ym−a| < δ which implies |f(xn) − f(ym)| < ϵ.  Thus lim n -> ∞ f(xn)   =  lim n -> ∞ f(yn).
QED.
ANOTHER WAY TO SHOW ALL SEQUENCES CONVERGE TO THE SAME LIMIT:
Interleave the two sequences x1, y1, x2, y2, x3, y3 forming sequence {zn}.
But lim n -> ∞ xn = a and lim n -> ∞ yn = a  so  lim n -> ∞ zn = a and similarly f(zn) converges to L say.
Then similarly f(xn) and f(yn) both converge to L.
