How do you solve #lim x-> pi/4# of #xcos^2x#?

Question

561 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-23T14:21:33

Substitute:

#lim_(x->pi/4) (xcos^2x)#

#=> pi/4cos^2(pi/4)#

#=>pi/4(1/sqrt(2))^2#

#=> pi/4(1/2)#

#=> (pi)/8#

Hopefully this helps!