Limit of a $\frac{4^{3x} \sin x}{3^{4x}}$ as $x\to\infty$

Question

$$\lim_{x \to \infty} \frac{4^{3x} \sin x}{3^{4x}}$$

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.

Answer 1

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

$$ \lim_{x \to \infty} \frac{1}{x} \cdot x \overset{\text{not true}}{=} \lim_{x \to \infty} 0 \cdot x = \lim_{x \to \infty} 0 = 0 $$

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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.

Answer 2

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.

Answer 3

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.