aggierabz2006zw

2022-07-09

Compute $\underset{x\to 0}{lim}\frac{{\mathrm{sin}}^{3}x}{\left(2x{\right)}^{3}}$

Expert

depending on the fact of $\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1$
$\frac{{\mathrm{sin}}^{3}}{8{x}^{3}}=\frac{1}{8}\frac{\mathrm{sin}x}{x}\frac{\mathrm{sin}x}{x}\frac{\mathrm{sin}x}{x}$

ziphumulegn

Expert

Recall that if f is continuous, then
$\underset{x\to a}{lim}f\left(g\left(x\right)\right)=f\left(\underset{x\to a}{lim}g\left(x\right)\right).$
Since the cube function $x↦{x}^{3}$ is continuous, we have
$\begin{array}{rl}\underset{x\to 0}{lim}\frac{{\mathrm{sin}}^{3}x}{\left(2x{\right)}^{3}}& ={\left(\frac{1}{2}\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}\right)}^{3}\\ & ={\left(\frac{1}{2}\right)}^{3}\\ & =\frac{1}{8}.\end{array}$

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