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If I am not mistaken, this expectation equals the coefficient of $x^{\varepsilon n}$ in $(1-x)^{\alpha n}(1+x)^{(1-\alpha) n}$ divided by the corresponding coefficient in $(1+x)^n$ (which equals, of course, $\binom{n}{\varepsilon n}$). Such coefficient may be represented as integrals over unit circle (and so over $[0,2\pi]$) and their asymptotics may be calculated by some standard machinery. I do not write down more, because I do not understand, what exactly do you want (say, which small constant is more, $\alpha$ or $\varepsilon$? Is $n$ chosen large after these constants are already fixed, or $n$ tends to infinity simultaneously with constants tending to zero?).