If $k$ is not a limit cardinal, then the cardinality of a set $E$ with $\lim\limits_{\alpha\in E}\alpha=k$ is equal to the cardinality of $k.$ In chapter 1.10 of the Tôhoku paper by Grothendieck, near the end of the proof of theorem 1.10.1, there is a statement adapted as follows.

Every set $E$ of ordinal numbers $\alpha\lt k$ with limit $\lim\limits_{\alpha\in E}\alpha= k$ has cardinality that of k.

Here $k$ is defined as the smallest ordinal number whose cardinality is strictly greater that the cardinality of some fixed set, say $F$ (I think if $\text{card }F=\aleph_{\lambda}$ then $k=\aleph_{\lambda+1}$).  
After some futile attempts to understand how the cardinality of $E$ can affect the limit, I have found nothing particularly helpful to show this statement.  
Thus any help or reference would be sincerely appreciated, thanks in advance.
 A: There is a concept of "regular cardinal" which means that there is no increasing function $f:\alpha\to\kappa$ with $\alpha<\kappa$ such that $\sup \{f(\beta):\beta<\alpha\}=\kappa$. It is known that every successor cardinal is regular. This is very standard and I think it would be easy to find a reference with this names. 
You can find it in the book Hrbacek-Jech: Introduction to set theory, or in http://www.math.unt.edu/~moliver/fa04s/JacksonNotes/card.pdf , Lemma 2.6 (using also Theorem 1.8))
In your case, since $\kappa$ is a successor cardinal (namely, $\aleph_{\alpha+1}$ for some ordinal $\alpha$), then it is regular, and if $E=\{\beta_i:i<\alpha\}$ is a subset of ordinals $\beta_i<\kappa$, also with $\alpha<\kappa$ then by defining 
\begin{align*}
f: &\alpha\longrightarrow \kappa\\
&i\longmapsto \beta_i
\end{align*}
you will contradict the fact that $\kappa$ is regular.
A: This is simply the statement that every successor cardinal is regular, meaning that it is equal to its own cofinality. It’s easiest to think about in terms of von Neumann cardinals, so that each cardinal is simply the set of smaller ordinals.
Let $\kappa=|E|$; then the smallest ordinal whose cardinality is greater than $\kappa$ is the ordinal (and cardinal) $\kappa^+$, the cardinal successor of $\kappa$. (This $\kappa^+$ is your $k$.) Suppose that $A\subseteq\kappa^+$, $\sup A=\kappa^+$, and $|A|<\kappa^+$. Then
$$\kappa^+=\bigcup A=\bigcup_{\alpha\in A}\alpha\;,$$
and $|\alpha|\le\alpha<\kappa^+$ for each $\alpha\in A$, so $|\alpha|\le\kappa$ for each $\alpha\in A$, and
$$\kappa^+=\left|\bigcup_{\alpha\in A}\alpha\right|\le\kappa\cdot|A|\;.$$
If $|A|<\kappa^+$, then $|A|\le\kappa$, and $\kappa\cdot|A|=\kappa<\kappa^+$, which is impossible, so we must have $|A|=\kappa^+$. (Clearly $|A|\le\kappa^+$, since $A\subseteq\kappa^+$.)
