# How do you evaluate \lim _ { x \rightarrow 0} \frac { \sin ^ { 3} 2x } { \sin ^ { 3} 3x }?

Apr 18, 2017

$\frac{8}{27}$

#### Explanation:

May I assume that you know the limit below:

${\lim}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

Let us now consider the limit in question.

${\lim}_{x \to 0} \frac{{\sin}^{3} 2 x}{{\sin}^{3} 3 x} = {\lim}_{x \to 0} {\left(\frac{\sin 2 x}{\sin 3 x}\right)}^{3}$

By squeeze the limit inside the parentheses,

$= {\left({\lim}_{x \to 0} \frac{\sin 2 x}{\sin 3 x}\right)}^{3}$

By dividing the numerator and the denominator by $x$,

$= {\left({\lim}_{x \to 0} \frac{\frac{\sin 2 x}{x}}{\frac{\sin 3 x}{x}}\right)}^{3}$

By dividing and multiplying the numerator by $2$ and the denominator by $3$,

$= {\left({\lim}_{x \to 0} \frac{2 \frac{\sin 2 x}{2 x}}{3 \frac{\sin 3 x}{3 x}}\right)}^{3}$

By applying the limit to the numerator and the denominator,

$= {\left(\frac{2 {\lim}_{x \to 0} \frac{\sin 2 x}{2 x}}{3 {\lim}_{x \to 0} \frac{\sin 3 x}{3 x}}\right)}^{3}$

Since $2 x \to 0$ and $3 x \to 0$ as $x \to 0$,

$= {\left(\frac{2 {\lim}_{2 x \to 0} \frac{\sin 2 x}{2 x}}{3 {\lim}_{3 x \to 0} \frac{\sin 3 x}{3 x}}\right)}^{3} = {\left(\frac{2 \cdot 1}{3 \cdot 1}\right)}^{3} = \frac{8}{27}$

I hope that this was clear.

Apr 18, 2017

$\frac{8}{27.}$

#### Explanation:

Knowing that,

$\sin 2 x = 2 \sin x \cos x ,$

and,

$\sin 3 x = 3 \sin x - 4 {\sin}^{3} x = \sin x \left(3 - 4 {\sin}^{2} x\right) ,$ we have,

$\text{The Limit=} {\lim}_{x \to 0} {\sin}^{3} \frac{2 x}{\sin} ^ 3 \left(3 x\right)$

$= {\lim}_{x \to 0} {\left(2 \sin x \cos x\right)}^{3} / {\left\{\sin x \left(3 - 4 {\sin}^{2} x\right)\right\}}^{3}$

$= {\lim}_{x \to 0} \frac{8 {\sin}^{3} x {\cos}^{3} x}{{\sin}^{3} x {\left(3 - 4 {\sin}^{2} x\right)}^{3}}$

$= {\lim}_{x \to 0} \frac{8 {\cos}^{3} x}{3 - 4 {\sin}^{2} x} ^ 3$

$= \frac{8 {\cos}^{3} 0}{3 - 4 {\sin}^{2} 0} ^ 3 = \frac{8 \cdot {1}^{3}}{3 - 4 \cdot {0}^{2}} ^ 3$

$\therefore \text{ The Limit=} \frac{8}{27.}$

Enjoy MAths.!