Let $X \sim B(n,c/n)$ be a binomially distributed random variable with parameter $p = c/n$, and hence mean $c$. Here $c$ is some function of $n$ such that

i) $c \geq n^{2/3}$

ii) The function $c$ grows slower than any linear function of $n$ (i.e., in big-O notation, $c = o(n)$, or equivalently $\lim_{n \to \infty} c/n = 0$).

For such a variable, I want a ball-park estimate of $E[X | X >= c]$, i.e., tail conditional expectation (TCE), for large $n$. If the probability $c/n$ were a constant, then by central limit theorem the TCE is approximately $c + \sqrt{c}$. However, $c/n$ is not a constant here. Also, I am most interested in only finding which of the following cases is true.

1. For all $c$ in the said range, the TCE is of the form $c + f(c)$ where $f(c)$ grows strictly slower than any linear function in c, i.e., $f(c) = o(c)$

2. There is some $c$ in the said range for which the TCE is of the form $c + f(c)$ where $f(c)$ is a linear function in $c$.

The choice of lower-bound for $c$, namely $c \geq n^{2/3}$ has no significance. I would be happy with resolving the question for a much more restricted range of $c$ by placing a bigger lower bound on $c$.

I tried writing the explicit expression for the TCE but I have not been able to get anything useful out of it. Also I saw a paper on TCE for binomial rv's, but it just gives the obvious formula obtained by using linearity of expectation and nothing more.