Let $X,Y$ be topological spaces, and $f:X\to Y$. We say that $f(x)\to l$ as $x \to b$ iff for every open neighborhood $N$ of $l$, there exists an open neighborhood $M$ of $b$ such that $f(M)\subset N$.
There is, however, a different weaker notion of limit: for any sequence $x_n$, whenever $x_n\to b$ as $n\to \infty$, then $f(x_n)\to l$ as $n\to \infty$.
I can prove that the first definition implies the second. In $\mathbb R$, the converse is also true. But is the converse also true in general topological spaces? I don't believe it is true but I cannot find counterexamples.
No.
The standard counterexample is to take $X=\omega_1+1$ with the order topology, $Y$ the two-point discrete space $\{0,1\}$, and $$ f\colon X\to Y;\quad \alpha\mapsto \begin{cases} 1 & \alpha=\omega_1\\ 0 & \text{otherwise} \end{cases}. $$ Then $f$ is sequentially continuous (since the only sequences that converge to $\omega_1$ are eventually $\omega_1$), but not continuous.