= Solution
Write $p=1+\varepsilon$ and assume $\lambda>0$. The <large-deviation speed> is $a_N=N^p$, and the answer is
$$
\boxed{I_A(x)=\begin{cases}\lambda x,&x\geq0,\\+\infty,&x<0.\end{cases}}
$$
Here and below a <large deviation principle> at speed $a_N$ means the open-set lower bound $\liminf a_N^{-1}\log\mathbb P(X_N\in G)\geq-\inf_G I$ and the closed-set upper bound $\limsup a_N^{-1}\log\mathbb P(X_N\in F)\leq-\inf_F I$. The displayed <rate function> is nontrivial and is a <good rate function>.
For $0<u<v$, the <exponential distribution> gives
$$
\mathbb P(A/a_N\in(u,v))=e^{-\lambda a_Nu}-e^{-\lambda a_Nv},
\qquad
\lim_N a_N^{-1}\log\mathbb P(A/a_N\in(u,v))=-\lambda u.
$$
Shrinking an interval about any $x>0$ supplies the local lower bound $-\lambda x$; neighborhoods of zero have <probability> tending to one. Every <open set> intersecting the nonnegative half-line contains one of these neighborhoods, giving the open-set bound. If a <closed set> $F$ contains zero, its upper bound is just $\mathbb P(\cdot)\leq1$. If it meets $[0,\infty)$ but excludes zero, closedness gives $b=\inf(F\cap[0,\infty))>0$, and $\mathbb P(A/a_N\in F)\leq e^{-\lambda a_Nb}$. A <closed set> disjoint from the support has <probability> zero. This proves the full <large deviation principle>, including the boundary at zero.
The fixed exponential tail explains the speed: an excursion of size $xN^{1+\varepsilon}$ costs $e^{-\lambda xN^{1+\varepsilon}}$. The full <Gärtner–Ellis theorem> is unnecessary here; the limiting <cumulant-generating function> is flat on $\theta<\lambda$ and infinite at $\theta\geq\lambda$, so it fails the theorem's <essential smoothness> hypothesis.
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