I think that these definitions of continuity are equivalent, but I'm not sure how to create a mathematical proof to convey this.
The first definition states that:
A function $f$ is continuous at a point $x=c$ if for every sequence ($a_n$)$_{n=1}^{\infty}$ that converges to $c$ such that ($a_n$) $\not=$ $c$ for all $n$, the sequence ($f$($a_n$))$_{n=1}^{\infty}$ converges to $f(c)$. If a function is continuous at every point of some set $S,$ we say that $f$ is continuous on $S.$
The second definition, the delta-epsilon definition, states that:
A function $f$ is said to be continuous at a point, $x$ if for every $\epsilon$>0 there exists a $\delta$>0 such that
|$f(y)-f(x)$|< $\epsilon$ for all $y$ such that |$y-x$|< $\delta$.
Hint 1:
If $f$ is delta-epsilon continuous and $(a_n)$ is such a sequence then given any $\varepsilon > 0$ you are looking for an $N$ such that if $n \ge N$ then
$$ |f(a_n) - f(c)| < \varepsilon. $$
Note that this looks similar to the inequality
$$ |f(x) - f(c)| < \varepsilon. $$
Hint 2:
For the converse direction, use the contrapositive. That is, assume $f$ is not continuous with the delta-epsilon definition. Write down what it means for $f$ not to satisfy the delta-epsilon definition without using the word "not" anywhere.
Now let $\delta_n = \frac{1}{n}$ and allow yourself to pick $a_n$ such that
$$ 0 < |a_n - c| < \delta_n \text{ but } |f(a_n) - f(c)| > \varepsilon_0 $$
where $\varepsilon_0 > 0$ is a fixed constant. Derive a contradiction.