Question #d54ed

Question

1858 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-11T20:33:52

Our goal is to rewrite this so that we can use

#lim_(thetararr0)sintheta/theta = lim_(thetararr0)theta/sintheta =1#

In this case, it looks like we must have #theta = 2x#

Note that

#(3x^2+2x)/(sin2x) = x/(sin2x)(3x+2)#

# = (2x)/(sin2x) sin(3x+2)/2#

Now we can evaluate the limit as #xrarr0#

#lim_(xrarr0) (3x^2+2x)/(sin2x) = lim_(xrarr0) (2x)/(sin2x) sin(3x+2)/2#

#lim_(xrarr0) (2x)/(sin2x) lim_(xrarr0) sin(3x+2)/2#

# = (1)(2/2)=1#