Limit of a $\frac{4^{3x} \sin x}{3^{4x}}$ as $x\to\infty$ $$\lim_{x \to \infty} \frac{4^{3x} \sin x}{3^{4x}}$$
I tried it in this way:\begin{align*}
&\Rightarrow \lim_{x \to \infty} \left(\frac{4^{3}}{3^{4}}\right)^x \sin x\\
&\Rightarrow \lim_{x \to \infty} \left(\frac{64}{81}\right)^x \sin x
\end{align*}
$\frac{64}{81}<1$ so we have $$\lim_{x \to \infty}(0)\times (\text{finite})=0$$
I am not sure if I did it in right way. This question came in $9$ marks I wonder why. Please correct me. 
 A: Perhaps the following reasoning is better:
\begin{align*}
\left|\left(\dfrac{64}{81}\right)^{x}\sin x\right|\leq\left(\dfrac{64}{81}\right)^{x}\rightarrow 0
\end{align*}
as $x\rightarrow\infty$, so by Squeeze Theorem the desired limit is zero as well.
A: Your reasoning is right but it's informal in the last step, unless you specifically have a theorem that $\lim 0 \times \text{finite} = 0$. You should probably cast that into a form where the Squeeze Theorem applies instead.
A: The form of your work suggests that you still have trouble understanding exactly what you're doing in that last step.
When you computed the limit of a product by plugging into the limits of the two factors, that's the actual value of the limit:
$$\begin{align*}
\lim_{x \to \infty} \left(\frac{64}{81}\right)^x \sin x
= 0 \cdot (\text{finite})
\end{align*} $$
Your work suggests you're still conceiving this as manipulating the quantities inside the limit, which is incorrect and often gets people in trouble; e.g. as in the following argument
$$ \lim_{x \to \infty} \frac{1}{x} \cdot x \overset{\text{not true}}{=} \lim_{x \to \infty} 0 \cdot x 
 = \lim_{x \to \infty} 0 = 0 $$

Additionally, while it is possible to make sense of $(\mathrm{finite})$ as being a "limit" of $\sin(x)$ as $x \to \infty$, it is a somewhat awkward sort of thing and probably not something you were actually taught how to reason with.
