$\phi(x)$ is bounded on $I_n=[\xi-\epsilon_n,\xi+\epsilon_n]$ and $\phi(\xi)\neq0$ ,where $\phi$ is continue at $\xi$, and $\epsilon_n=o(1)$ . $h(x)\in C^2$ is continous ,and $\xi$ is the only maximum point of $h(x)$.
How does this come ?: $$\int_{I_n}\phi(x)e^{n(h(x)-h(\xi))}dx$$
$$=(1+o(1))\phi(\xi)\int_{I_n}e^{n(h(x)-h(\xi))}dx$$ as $n\rightarrow \infty$
Let $M_n=\max_{I_n} \phi(x)$ and $m_n=\min_{I_n} \phi(x)$. Then for the integral
$$ m_n \int_{I_n} e^{n(h(x)-h(\xi))} dx \leq \int_{I_n} \phi(x) e^{n(h(x)-h(\xi)} dx \leq M_n \int_{I_n} e^{n(h(x)-h(\xi))} dx $$ and $$ m_n\int_{I_n} e^{n(h(x)-h(\xi))} dx \leq \phi(\xi) \int_{I_n} e^{n(h(x)-h(\xi)} dx \le M_n \int_{I_n} e^{n(h(x)-h(\xi))} dx $$ but since $(M_n-m_n)= \phi(\xi)+o(1)$ due to the continuity of $\phi$, we get it.