Suppose that $X_1,X_2,\ldots, X_n$ are i.i.d random variables with continuous density $f(x)$, which is defined in the whole $\mathbb{R}$. Consider $$s(x)=\lim_{n\to\infty}\frac{1}{n}\log\mathbb{P}(\overline{X}>x)$$ and $T(x)=\log\mathbb{P}(X_1>x)$ Here $\overline{X}=\frac{X_1+\ldots+X_n}{n}$.
Under what assumptions is true that $$\lim_{x\to\infty}\frac{s(x)}{T(x)}=1\text{ ?}$$ For example, this holds for the Gaussian, the exponential, the symmetric exponential.
$\newcommand{\R}{\mathbb R}$
Proposition 1: For \begin{equation*} s(x)\sim T(x)=\ln P(X>x) \tag{00}\label{00} \end{equation*} to hold (as $x\to\infty$), it is necessary and sufficient that \begin{equation*} T(x)\sim T_0(x) \tag{10}\label{10} \end{equation*} for some real-valued concave function $T_0$.
The relation $a\sim b$ is understood here as $a/b\to1$. In particular, $a\sim b$ implies that eventually the values of $|b|$ are not in the set $\{0,\infty\}$.
Proof of Proposition 1:
Part 1: Necessity: Assume that \eqref{00} holds; in particular, this implies that $T(x)\ne-\infty$ eventually (that is, for all large enough $x>0$); it is clear that $T(x)\ne0$ eventually. We have to show that then \eqref{10} holds as well.
By Cram´er's large deviation principle (see e.g. Corollary 2.2.19), for all real $x$ \begin{equation*} s(x)=-\inf_{y\in[x,\infty)}L^*(y), \tag{20}\label{20} \end{equation*} where \begin{equation*} L^*(y):=\sup_{t\in\R}(ty-L(t)),\quad L(t):=\ln Ee^{tX}, \tag{30}\label{30} \end{equation*} $X:=X_1$.
If $L(t)=\infty$ for all real $t>0$, then $L^*(y)=0$ for all real $y\ge0$, and hence, by \eqref{20}, $s(x)=0$ for all real $x\ge0$, so that \eqref{00} will not hold.
So, without loss of generality (wlog) \begin{equation} \text{$L(t_0)<\infty$ for some real $t_0>0$, } \tag{40}\label{40} \end{equation} and then, by \eqref{30}, $L^*(y)\ge t_0y-L(t_0)\to\infty$ as $y\to\infty$. Also, the function $L^*$ is convex, being the supremum of affine functions. So, by \eqref{20}, for all large enough real $x>0$ \begin{equation*} s(x)=-L^*(x)=\ln\inf_{t\in\R}Q(x,t), \tag{20a}\label{20a} \end{equation*} where \begin{equation*} Q(x,t):=e^{-tx}M(t),\quad M(t):=Ee^{tX}. \tag{50}\label{50} \end{equation*} Letting $T_0:=-L$ in a neighborhood of $\infty$ and looking at the first equality in \eqref{20a}, we see that \eqref{00} does imply \eqref{10}, for such a real-valued concave function $T_0$, which completes the necessity part of the proof.
Part 2: Sufficiency: Assume that \eqref{10} holds for some real-valued concave function $T_0$. We have to show that then \eqref{00} holds as well.
Since $T(x)\to-\infty$ (as $x\to\infty$), \eqref{10} and the concavity of $T_0$ imply that for some $\tau\in[-\infty,0)$ we have $T_0(x)/x\to\tau$ and $T(x)/x\to\tau$. So, \eqref{40} and \eqref{20a} hold.
It also follows that there exists $\mu:=EX\in[-\infty,\infty)$. So, by Jensen's inequality, for $x>\mu$ and $t<0$ we have $Q(x,t)\ge e^{t(\mu-x)}>1=Q(x,0)$. So, for all large enough real $x$ \begin{equation*} s(x)=-L^*(x)=\ln\inf_{t\ge0}Q(x,t). \tag{20b}\label{20b} \end{equation*}
Let \begin{equation*} q(x):=P(X>x), \end{equation*} so that \begin{equation*} T(x)=\ln q(x). \tag{60}\label{60} \end{equation*} By Markov's inequality, $q(x)\le Q(x,t)$ for all real $t\ge0$. So, by \eqref{20b}, \begin{equation*} s(x)\ge T(x) \tag{70}\label{70} \end{equation*} eventually (that is, for all large enough real $x>0$).
By \eqref{50}, for real $t>0$,
\begin{equation*}
\begin{aligned}
M(t)=Ee^{tX}&=-\int_{-\infty}^\infty dq(u)e^{tu} \\
&=-\int_{-\infty}^\infty dq(u)\int_{-\infty}^u t\,dv\,e^{tv} \\
&=-\int_{-\infty}^\infty t\,dv\,e^{tv}\int_v^\infty dq(u) \\
&=t\int_{-\infty}^\infty \,dv\,e^{tv} q(v) \\
&=I_1+I_2,
\end{aligned}
\tag{80}\label{80}
\end{equation*}
where
\begin{equation*}
I_1:=t\int_{-\infty}^{x-A} \,dv\,e^{tv} q(v)
\le t\int_{-\infty}^{x-A} \,dv\,e^{tv}=e^{t(x-A)}, \tag{90}\label{90}
\end{equation*}
\begin{equation*}
I_2:=t\int_{x-A}^\infty \,dv\,e^{tv} e^{-g(v)}, \tag{100}\label{100}
\end{equation*}
$A$ is a real number (to be specified later), and
$$g(v):=-\ln q(v)=-T(v).$$
Let
$$g_0(v):=-T_0(v).$$
The function $g_0$ is convex, (strictly) increasing, $>0$, and $<\infty$ in a neighborhood of $\infty$ -- say on the interval \begin{equation*} [a,\infty) \end{equation*} for some real $a$. So, $g_0'>0$ on $[a,\infty)$, where $g_0'$ is (say) the right derivative of the convex function $g_0$. In what follows, by default $x\in[a,\infty)$.
Take now any $c\in(0,1)$ and let
\begin{equation*}
A:=A(x):=\frac{cg_0(x)}{g_0'(x)}, \tag{110}\label{110}
\end{equation*}
so that eventually $A>0$.
Note also that, by the convexity of $g_0$ on $[a,\infty)$, we have $(x-a)g_0'(x)\ge g_0(x)-g_0(a)$ (for $x\in[a,\infty)$), that is, $xg_0'(x)-g_0(x)\ge ag_0'(x)-g_0(a)$, so that
\begin{equation*}
x-A=(1-c)x+c\frac{xg_0'(x)-g_0(x)}{g_0'(x)} \\
\ge(1-c)x+c\frac{ag_0'(x)-g_0(a)}{g_0'(x)}. \tag{120}\label{120}
\end{equation*}
Also, again by the convexity of $g_0$ on $[a,\infty)$, there exists $g_0(\infty):=\lim_{x\to\infty}g_0'(x)\in(0,\infty]$. So, by \eqref{120}, $x-A\to\infty$ and hence wlog $x-A\ge a$, which will be assumed by default in the sequel.
Let now \begin{equation*} k(x):=\inf_{v\in[x-A,\infty)}\frac{g(v)}{g_0(v)}, \tag{130}\label{130} \end{equation*} so that, in view of \eqref{10} and because $x-A\to\infty$, we have \begin{equation*} k(x)\to1. \tag{140}\label{140} \end{equation*}
Now comes the crucial point, which is choosing the value of $t$ as follows: \begin{equation*} t:=t(x):=k(x)g_0'(x)-\frac1A=\frac{ck(x)g_0(x)-1}A. \tag{150}\label{150} \end{equation*}
Using the convexity of $g_0$ on $[a,\infty)$ again, we have
\begin{equation*}
g_0(v)\ge h_x(v):=g_0(x)+g_0'(x)(v-x)
\end{equation*}
for real $v\ge x-A$. So,
by \eqref{100}, \eqref{130}, \eqref{150}, \eqref{140}, and \eqref{10},
\begin{equation*}
e^{-tx}I_2\le \int_{x-A}^\infty \,dv\,e^{tv} e^{-k(x)h_x(v)} \\
=e^{-k(x)g_0(x)+1}(ck(x)g_0(x)-1)=e^{-k(x)g_0(x)(1+o(1))}=q(x)^{1+o(1)} \tag{160}\label{160}
\end{equation*}
(as $x\to\infty$).
Also, by \eqref{90} and \eqref{150},
\begin{equation*}
e^{-tx}I_1\le e^{-tA}=e^{-ck(x)g_0(x)+1}=eq(x)^{ck(x)}.
\end{equation*}
In view of \eqref{140} and because $c\in(0,1)$ was arbitrary, we now get
\begin{equation*}
e^{-tx}I_1\le q(x)^{1+o(1)}. \tag{170}\label{170}
\end{equation*}
It follows from \eqref{20b}, \eqref{50}, \eqref{80}, \eqref{170}, \eqref{160}, and \eqref{60} that
\begin{equation*}
s(x)\le\ln\big(q(x)^{1+o(1)}\big)\sim\ln q(x)=T(x). \tag{180}\label{180}
\end{equation*}
Finally, \eqref{00} follows from \eqref{70} and \eqref{180}.
This completes the sufficiency part of the proof as well. $\quad\Box$
