Compute $\lim \limits _{x\to 0} \frac{\sin^3x}{(2x)^3}$ How can I compute $\lim \limits _{x\to 0} \frac{\sin^3x}{(2x)^3}$?
 A: depending on the fact of $\lim \limits _{x\to 0} \frac{\sin x}{x}=1$
$$\frac{\sin^3}{8x^3}=\frac{1}{8}\frac{\sin x}{x}\frac{\sin x}{x}\frac{\sin x}{x}$$
A: You can use the L'Hospital's rule.
Take derivative of both the numerator and the denominator until they are not zeroes. In your case, take the derivative 3 times, and your denominator is no long zero.
A: $$\lim_{x\to 0} \frac{\sin^3(x)}{(2x)^3}=$$
$$\lim_{x\to 0} \frac{\sin^3(x)}{8x^3}=$$
$$\frac{1}{8}\lim_{x\to 0} \frac{\sin^3(x)}{x^3}=$$
$$\frac{1}{8}\lim_{x\to 0} \frac{\frac{\text{d}}{\text{d}x}\sin^3(x)}{\frac{\text{d}}{\text{d}x}x^3}=$$
$$\frac{1}{8}\lim_{x\to 0} \frac{3\sin^2(x)\cos(x)}{3x^2}=$$
$$\frac{1}{8}\lim_{x\to 0} \frac{\sin^2(x)\cos(x)}{x^2}=$$
$$\frac{\left(\lim_{x\to 0}\cos(x)\right)\left(\lim_{x\to 0}\frac{\sin^2(x)}{x^2}\right)}{8}=$$
$$\frac{\left(\cos(0)\right)\left(\lim_{x\to 0}\frac{\sin^2(x)}{x^2}\right)}{8}=$$
$$\frac{\left(1\right)\left(\lim_{x\to 0}\frac{\sin^2(x)}{x^2}\right)}{8}=$$
$$\frac{1}{8}\lim_{x\to 0}\frac{\sin^2(x)}{x^2}=$$
$$\frac{1}{8}\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\sin^2(x)}{\frac{\text{d}}{\text{d}x}x^2}=$$
$$\frac{1}{8}\lim_{x\to 0}\frac{2\sin(x)\cos(x)}{2x}=$$
$$\frac{1}{8}\lim_{x\to 0}\frac{\sin(x)\cos(x)}{x}=$$
$$\frac{\left(\lim_{x\to 0}\cos(x)\right)\left(\lim_{x\to 0}\frac{\sin(x)}{x}\right)}{8}=$$
$$\frac{\left(\cos(0)\right)\left(\lim_{x\to 0}\frac{\sin(x)}{x}\right)}{8}=$$
$$\frac{\left(1\right)\left(\lim_{x\to 0}\frac{\sin(x)}{x}\right)}{8}=$$
$$\frac{\lim_{x\to 0}\frac{\sin(x)}{x}}{8}=$$
$$\frac{1}{8}\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\sin(x)}{\frac{\text{d}}{\text{d}x}x}=$$
$$\frac{1}{8}\lim_{x\to 0}\frac{\cos(x)}{1}=$$
$$\frac{1}{8}\lim_{x\to 0}\cos(x)=$$
$$\frac{1}{8}\cdot\cos(0)=$$
$$\frac{1}{8}\cdot 1=\frac{1}{8}$$
A: Here is an approach that does not rely on either L'Hospital's Rule or series expansions.  We simply recall from geometry the inequalities 
$$|x\cos x|\le |\sin x| \le |x| \tag 1$$
for $-\pi/2\le x\le \pi/2$.  Using $(1)$, we obtain
$$\left|\frac{\cos x}{2}\right|^3\le \left|\frac{\sin x}{2x}\right|^3\le \frac18$$
Therefore, by the squeeze theorem, we find that 
$$\lim_{x\to 0}\left(\frac{\sin x}{2x}\right)^3=\frac18$$
A: 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*}
