find the limit to infinity $a_n= 3^n ~\sin(4^{-n})~$

Question

I am having problems solving limits when its infinity multiplied by $~0~$

$$\lim_{n\to \infty} a_n= 3^n \sin (4^{-n})~$$

I can't use L'Hôpital's rule here so how do I solve this

Answer 1

We have $a_n=3^n \sin(\frac{1}{4^n}).$

Hint: $| \sin(x)| \le |x|$ for all $x$.

Can you proceed ?

Answer 2

$$3^n\sin 4^{-n}=\left(\frac{3}{4}\right)^n\cdot\frac{\sin{4^{-n}}}{4^{-n}}\rightarrow0\cdot1=0.$$

Answer 3

Simpler with equivalents:

We know that $\sin x\sim_0x$, so $\sin 4^{_n}\sim_\infty 4{-n}$, hence $$3^n\sin 4^{-n}\sim_\infty\Bigl(\frac34\Bigr)^n,\quad\text{which tends to }0. $$