$lim_{i \to \infty}e_i$ is a sequence which is not convergent, where $e_i$ denote the sequence whose $i$th term is 1 and rest all are zero. From this point of view, we can say $\lim_{i \to \infty}e_i\notin c_0$. But we know $\lim_{i \to \infty}e_i\in c_0$. Where I am wrong.
($c_0$ is a normed space of the sequences which converge to $0$ and is equipped with supremum norm.)
You should be very careful of which topology you are working with.
Your sequence $(e_i)$, which I presume to mean $e_i = (\delta_{in} : n\geq1)$, does not converge under the norm topology on $c_0(\mathbb{N})$. This is easily checked by noticing that $\|e_i - e_j\| = 1$ whenever $i \neq j$.
On the other hand, it converges to $0$ under the weak topology on $c_0(\mathbb{N})$. That is, we have $\ell(e_i) \to 0$ for any bounded linear functional $\ell$ on $c_0(\mathbb{N})$. (If you want to verify this by yourself, a hint is that the dual space $c_0(\mathbb{N})^*$ can be identified as the space $\ell^1(\mathbb{N})$ of absolutely summable sequences.)
This also implies that $e_i$ converges pointwise to $0$. That is, we have $(e_i)_n = \delta_{in} \to 0$ as $i\to\infty$ for each fixed $n$.
And these 3 topologies (norm topology, weak topology, topology induced from the product space $\mathbb{R}^{\mathbb{N}}$) are all different!