Question #d54ed
1 Answer
Mar 11, 2017
Our goal is to rewrite this so that we can use
In this case, it looks like we must have
Note that
#(3x^2+2x)/(sin2x) = x/(sin2x)(3x+2)#
# = (2x)/(sin2x) sin(3x+2)/2#
Now we can evaluate the limit as
#lim_(xrarr0) (2x)/(sin2x) lim_(xrarr0) sin(3x+2)/2#
# = (1)(2/2)=1#