Question

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

Answer 1

`8/27`

Explanation:

May I assume that you know the limit below:

`lim_(theta to 0)(sin theta)/theta=1`

Let us now consider the limit in question.

`lim_(x to 0)(sin^3 2x)/(sin^3 3x)=lim_(x to 0)((sin2x)/(sin3x))^3`

By squeeze the limit inside the parentheses,

`=(lim_(x to 0)(sin2x)/(sin3x))^3`

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

`=(lim_(x to 0)((sin2x)/x)/((sin3x)/x))^3`

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

`=(lim_(x to 0)(2(sin2x)/(2x))/(3(sin3x)/(3x)))^3`

By applying the limit to the numerator and the denominator,

`=((2lim_(x to 0)(sin2x)/(2x))/(3lim_(x to 0)(sin3x)/(3x)))^3`

Since `2x to 0` and `3x to 0` as `x to 0`,

`=((2lim_(2x to 0)(sin2x)/(2x))/(3lim_(3x to 0)(sin3x)/(3x)))^3=((2cdot1)/(3cdot1))^3=8/27`

I hope that this was clear.

Answer 2

`8/27.`

Explanation:

Knowing that,

`sin2x=2sinxcosx,`

and,

`sin3x=3sinx-4sin^3x=sinx(3-4sin^2x),` we have,

`"The Limit="lim_(x to 0) sin^3(2x)/sin^3(3x)`

`=lim_(x to 0) (2sinxcosx)^3/{sinx(3-4sin^2x)}^3`

`=lim_(x to 0) (8sin^3xcos^3x)/{sin^3x(3-4sin^2x)^3}`

`=lim_(x to 0)(8cos^3x)/(3-4sin^2x)^3`

`={8cos^3 0)/(3-4sin^2 0)^3=(8*1^3)/(3-4*0^2)^3`

`:." The Limit="8/27.`

Enjoy MAths.!