Let $S_n$ be a sequence of positive random variables, such that the following tail bound holds for any $n$: $$ \mathbb{P} (S_n \geq t) \leq e^{- n t} \hspace{1cm} \text{for sufficiently large $t$} $$ I want to show that $\lim_{n \to \infty} \mathbb{E} [ f(S_n) \mid S_n \geq t] \mathbb{P}(S_n \geq t) = 0$, where $f$ is a polynomial function. If $f$ is identity, then it's easy to show that, but for general I don't know if there is a straight way or not.
So far, I tried the following approach which seems it works: $$ \mathbb{E} [ f(S_n) \mid S_n \geq t] \mathbb{P}(S_n \geq t) = \int_t^{\infty} f(x) \mathbb{P}(x) dx \leq \int_t^{\infty} f(x) \mathbb{P}(S_n \geq x) dx \leq \int_t^{\infty} f(x) e^{-nx} dx $$ then by Laplace method, we can approximate this integral for large $n$, and show that it goes to zero as $n \to \infty$.
I wonder if there is an easier/ more straightforward approach to show this.
It suffices to deal with the case $f(x)=x^k$ for some $k$, by linearity of expectation. Notice that $$ \mathbb{E} [ f(S_n) \mid S_n \geq t] \mathbb{P}(S_n \geq t) = \mathbb{E} \left[ f(S_n) \mathbf{1}_{\{S_n \geq t\}} \right]=\mathbb{E} \left[ S_n^k \mathbf{1}_{\{S_n \geq t\}} \right] $$ hence $$ \mathbb{E} [ f(S_n) \mid S_n \geq t] \mathbb{P}(S_n \geq t) =k\int_0^\infty s^{k-1}\mathbb P\left(S_n\geq s,S_n\geq t\right)ds =k\int_0^ts^{k-1}ds\mathbb P(S_n\geq t)+k\int_t^\infty s^{k-1}\mathbb P\left(S_n\geq s\right)ds. $$ Then use the given bound on the tail of $S_n$.