In general we say that a function $f\left(n\right)$ satisfies a large deviation principle if: \begin{equation} \lim_{n\rightarrow\infty}-\frac{1}{n}\ln \left[f\left(n\right)\right]=F \end{equation} What we really want to mean for practical purposes is that: \begin{equation} f\left(n\right)\approx e^{-nF} \end{equation} but in textbooks it is found as: \begin{equation} f\left(n\right)\asymp e^{-nF} \end{equation} where the symbol "$\asymp$" means: \begin{equation} a_{n}\asymp b_{n}\Leftrightarrow\lim_{n\rightarrow\infty}\frac{1}{n}\ln a_{n}=\lim_{n\rightarrow\infty}\frac{1}{n}\ln b_{n} \end{equation} Now my question is:
How can I go from the first equation to the third one?
Just to mention, in general it should be true that: \begin{equation} \lim_{n\rightarrow\infty}-\frac{1}{n}\ln \left[f\left(n\right)\right]\neq -\frac{\lim_{n\rightarrow\infty}\ln\left[f\left(n\right)\right]}{\lim_{n\rightarrow\infty}n} \end{equation} right?
Suppose the first equation is fulfilled. To prove the third one, we have to check whether $$ \def\li{\lim_{n\to\infty}\frac 1n \log}\li f(n) = \li e^{-nF}\tag 1 $$ But $$ \li e^{-nF} = \lim_{n\to\infty}\frac 1n \cdot (-nF) = -F $$ hence (1) reads $$ \li f(n) = -F $$ and that is true by your first equation.