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

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.

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.!