find the limit to infinity $a_n= 3^n ~\sin(4^{-n})~$ 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
We have $a_n=3^n \sin(\frac{1}{4^n}).$

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

Can you proceed ?
Alternatively, consider the series:
$$\sum_{n=1}^\infty 3^n \sin (4^{-n})$$
By the ratio test:
$$\lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\lim_{n\to \infty} \frac{3^{n+1}\sin (4^{-(n+1)})}{3^n\sin (4^{-n})}=
\lim_{n\to \infty} \frac{3\sin(4^{-(n+1)})}{\sin(4\cdot 4^{-(n+1)})}=\\
\lim_{n\to \infty} \frac{3\require{cancel}\cancel{\sin(4^{-(n+1)})}}{4\require{cancel}\cancel{\sin (4^{-(n+1)})}\cos{(4^{-(n+1)})}\cos(2(4^{-(n+1)}))}=\frac34&lt;1.$$
The convergence of the series implies the limit of the general term is $0$.