= Solution
For the <uniform distribution> on $[0,1]$, the mean is $1/2$ and the <variance> is $1/12$. The <central limit theorem> for independent identically distributed variables with finite nonzero <variance> gives
$$
Z_n=\frac{S_n-n/2}{\sqrt{n/12}}\ \xrightarrow{d}\ N(0,1).
$$
Choose $\boxed{\alpha=1/2,\quad\beta=1/2}$. Then
$$
\mathbb P(S_n<\alpha n+n^\beta)=\mathbb P(Z_n<\sqrt{12})\longrightarrow\boxed{\Phi(\sqrt{12})}.
$$
The standard normal distribution function is continuous and strictly increasing, so its value at the finite positive number $\sqrt{12}$ is strictly between $1/2$ and one, as required.
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