= Solution
The finite-energy comparison $I_\varepsilon(u_\varepsilon)\leq I_\varepsilon(x)<\infty$ gives $\varepsilon\int|u_\varepsilon'|^{4/3}<\infty$. Together with the endpoint value and <absolute continuity>, this proves $u_\varepsilon\in W^{1,4/3}(0,1)$.
For $\alpha>1/4$, the power competitor $u_\alpha=x^\alpha$ is admissible and has finite <derivative> energy. Direct integration gives
$$
I_\varepsilon(u_\alpha)=\frac1{8\alpha+1}-\frac2{4\alpha+2}+\frac13+\frac{3\varepsilon\alpha^{4/3}}{4\alpha-1}.
$$
Choose $\delta=\varepsilon^{1/3}$ and $\alpha=1/4+\delta$. The first three terms tend to zero as $\alpha\downarrow1/4$, while the last equals $\tfrac34\varepsilon^{2/3}(1/4+\varepsilon^{1/3})^{4/3}\to0$. By minimality and nonnegativity,
$$
\boxed{0\leq I_\varepsilon(u_\varepsilon)\leq I_\varepsilon(u_{1/4+\varepsilon^{1/3}})\longrightarrow0.}
$$
One also obtains the quantitative bound $I_\varepsilon(u_\varepsilon)=O(\varepsilon^{2/3})$: for $\alpha=1/4+\delta$, the <mean value theorem> gives $|x^{4\alpha}-x|\leq4\delta x|\log x|$, and $\int_0^1x^2\log^2x\,dx=2/27$. Thus the squared mismatch is at most $32\delta^2/27$, matching the order of the <derivative> penalty with this choice. This is a <power-law comparison for vanishing derivative regularization>.
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