<s>[INST] Compute $\lim \limits _{x\to 0} \frac{\sin^3x}{(2x)^3}$ How can I compute $\lim \limits _{x\to 0} \frac{\sin^3x}{(2x)^3}$?
 [/INST] Recall that if $f$ is continuous, then
$$\lim_{x\rightarrow a}f(g(x))=f\left(\lim_{x\rightarrow a}g(x)\right).$$
Since the cube function $x\mapsto x^3$ is continuous, we have
\begin{align*}
\lim_{x\rightarrow 0}\frac{\sin^3 x}{(2x)^3} &= \left(\frac{1}{2}\lim_{x\rightarrow 0}\frac{\sin x}{x}\right)^3\\
&= \left(\frac{1}{2}\right)^3\\
&= \frac{1}{8}.
\end{align*}
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