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?

$$\lim_{n \rightarrow\infty}\left(\frac{3^{-n}\sin(3^{(1-n)})}{\tan(3^{1-2n})} \right)$$

Putting $h=3^{-n}$

As $n\to \infty \implies h\to 0$

and $3^{1-n}=3\cdot3^{-n}=3h ; 3^{1-2n}=3\cdot(3^{-n})^2=3h^2$

$$\lim_{n \rightarrow\infty}\left(\frac{3^{-n}\sin(3^{(1-n)})}{\tan(3^{1-2n})} \right)$$

$$=\lim_{h \rightarrow0}\left(\frac{h \sin(3h)}{\tan(3h^2)} \right)$$

$$=\left(\lim_{h \rightarrow0}\frac{\sin3h}{3h}\right)\cdot\left(\lim_{h \rightarrow0}\frac{3h^2}{\sin3h^2}\right)\cdot \left(\lim_{h \rightarrow0}\cos(3h^2)\right)$$

$$=1$$

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 $$\displaystyle \lim_{x \to 0} \dfrac{\sin(x)}x = 1 = \displaystyle \lim_{y \to 0} \dfrac{\tan(y)}y$$

Hint: Use this fact that if $\alpha(x)\sim 0$ (I mean it is very small) when $x\to 0$, then $$\sin(\alpha(x))\sim \alpha(x)$$ and $$\tan(\alpha(x))\sim \alpha(x)$$

• $\sim$ comes in handy time and time again! $\;+1\quad \ddot\smile\quad$ – Namaste Mar 20 '13 at 19:50