What is #lim_(x to pi) (sin3x)/(sin2x) #?

3 Answers
Jan 21, 2018

#-3/2#

Explanation:

#lim_(x->pi)(sin3x)/(sin2x)-> 0/0#

So we can use L'Hopital's rule:

#d/dxsin3x = 3cos3x#

#d/dxsin2x=2cos2x#

#thereforelim_(x->pi)(sin3x)/(sin2x)=lim_(x-> pi)(3cos3x)/(2cos2x)#

#=(3cos(3pi))/(2cos(2pi))=(3*(-1))/(2*1)=-3/2#

This is in agreement with the graph of the function:

graph{(sin(3x))/(sin(2x)) [-7.31, 12.69, -5.16, 4.84]}

Clicking on the graph at the right location, it can be seen that at #x=pi#, #y=-3/2=-1.5#

Jan 21, 2018

#lim_(x to pi) (sin3x)/(sin2x)=-3/2#

Explanation:

#lim_(x to pi) (sin3x)/(sin2x)=_(D.L.H)^((0/0))#

#lim_(x to pi) ((3x)'cos3x)/((2x)'cos2x)# #=#

#3/2lim_(x to pi) (cos3x)/(cos2x)# #=#

#3/2*(cos3pi)/(cos2pi)# #=#

#-3/2#

because
#cos3pi=-1#
#cos2pi=1#
enter image source here

Jan 21, 2018

#lim_(x->pi)(sin3x)/(sin2x)=-3/2#

Explanation:

Here's a method without using l'Hopital's rule. I personally feel this method to be more challenging and fun.

#sin3x=sin(2x+x)=sin2xcosx+cos2xsinx=#

#2sinxcos^2x+(cos^2x-sin^2x)sinx=#

#2sinxcos^2x+cos^2xsinx-sin^3x=#

#sinx(2cos^2x-sin^2x+cos^2x)=#

#sinx(2cos^2x-sin^2x+1-sin^2x)=#

#sinx(2cos^2x-2sin^2x+1)=#

#sinx(2cos2x+1)#

#sin2x=2sinxcosx#

So #(sin3x)/(sin2x)=(sinx(2cos2x+1))/(2sinxcosx)=(2cos2x+1)/(2cosx)#

And #lim_(x->pi)(sin3x)/(sin2x)=lim_(x->pi)(2cos2x+1)/(2cosx)=(2cos2pi+1)/(2cospi)=-3/2#