Calculate: $\lim_{n \rightarrow\infty}\left(\frac{3^{-n}\sin(3^{(1-n)})}{\tan(3^{1-2n})} \right)$ How to calculate following limit without using L'Hospital rule?
$$\lim_{n \rightarrow\infty}\left(\frac{3^{-n}\sin(3^{(1-n)})}{\tan(3^{1-2n})} \right)$$

Rewrite the expression as $$\dfrac{3^{-n}\sin(3^{1-n})}{\tan(3^{1-2n})} = \dfrac{\sin(3^{1-n})}{3^{1-n}} \cdot \dfrac{3^{1-2n}}{\tan(3^{1-2n})}$$
and note that $3^{1-n} \to 0$ and $3^{1-2n} \to 0$ as $n \to \infty$. Now make use of the fact that $$\displaystyle \lim_{x \to 0} \dfrac{\sin(x)}x = 1 = \displaystyle \lim_{y \to 0} \dfrac{\tan(y)}y$$
