lim as x approaches 0
(1-cos(2x))/(3sinx)
How do I derive both the numerator and denominator
using L'Hopital's rule
lim (1-cos(2x))/(3sinx) as x ---> 0
= lim 2sin2x/3cosx
= 4sinxcosx/3cosx
= (4/3)sinx , as x --->
= 0
How did you get 4sinxcosx
sin(2A) = 2sinA cosA, one of the standard identities.
So that's saying sin(2x)=2sinxcosx?
What if there was no 2 in sin(2x)? I dont rememeber the identities is there a link to all the identities somewhere
have you seen
sin(A+B) = sinAcosB + cosAsinB ???
then sin (2A) = sin(A+A)
= sinAcosA + cosAsinA
= 2sinAcosA, called the double angle formula.
No we haven't went over that identity in my class. So I was wondering if there was another way to solve this question
could just go from:
lim 2sin2x/3cosx as x----> 0
= 2(0)/(3(1))
= 0/3
= 0