Find the limit of the following sequence or determine that the sequence diverges. \( a_{n}=\frac{e^{-n}}{10 \sin \left(e^{-n}\right)} \)

To find the limit of the sequence \( a_n = \frac{e^{-n}}{10 \sin(e^{-n})} \) as \( n \) approaches infinity, we first analyze the behavior of \( e^{-n} \) as \( n \to \infty \). Notice that as \( n \to \infty \), \( e^{-n} \) approaches 0. Consequently, we can substitute \( x = e^{-n} \), leading to \( x \to 0 \) as \( n \to \infty \). We can then rewrite the expression in terms of \( x \): \[ a_n = \frac{x}{10 \sin(x)}. \] Now let's analyze \( \sin(x) \) using the fact that \( \sin(x) \approx x \) when \( x \) is close to 0. Therefore, we can approximate \( \sin(e^{-n}) \) as follows: \[ \sin(e^{-n}) \sim e^{-n} \quad \text{(as \( n \to \infty \))}. \] Substituting this approximation into our expression for \( a_n \): \[ a_n \approx \frac{e^{-n}}{10 e^{-n}} = \frac{1}{10}. \] Now, let's rigorously evaluate the limit: Using the small-angle approximation for sine, we have: \[ \sin(e^{-n}) = e^{-n} - \frac{(e^{-n})^3}{6} + O((e^{-n})^5). \] Thus, as \( n \to \infty \): \[ \sin(e^{-n}) \sim e^{-n}. \] This leads us back to our limit: \[ a_n = \frac{e^{-n}}{10 \sin(e^{-n})} \rightarrow \frac{e^{-n}}{10 e^{-n}} = \frac{1}{10}. \] Therefore, the limit of the sequence \( a_n \) as \( n \to \infty \) is \[ \lim_{n \to \infty} a_n = \frac{1}{10}. \] Thus, we conclude: \[ \boxed{\frac{1}{10}}. \]