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*** I am assuming that by (cos 2x - 1) you mean [cos(2x) - 1] . It's an indeterminate form 0/0 , so you can use L'Hopital's rule (differentiate numerator and denominator separately). [ cos(2x) - 1 ] / [ 1 - cos(3x) ] lim x -> 0 = [ - 2sin(2x) ] / 3sin(3x) lim x -> 0 That is also an indeterminate form 0/0 , so apply L'Hopital's rule again : = [ - 4cos(2x) ] / 9cos(3x) lim x -> 0 = (- 4 / 9) ------- Here's a zoomed-in portion of the graph of f(x) = [ cos(2x) - 1 ] / [ 1 - cos(3x) ] You can see the graph passing through the point ( 0 , -4/9 )
were did 3x come from
@morganKING i am getting -1 for the limit of this one, but when i graph i am getting f(t)=3, as t approaches 0
Are you having trouble finding the derivative? That is the only scenario I can see here since it tells you to use lhospital.
let f(t)=t(1-cos(t)) and g(t)=t-sin(t) f(0)=0 and g(0)=0 since we have 0/0 we can use l'hopital rule (i think it is much easier to multiply the bottom by 1+sin(t) by whateves; it doesn't say to do that) Do limt->0 (f'/g') since we have f/g=0/0
is the answer 1?
Not for your problem you asked about.
yeah the derivative confused me @myininaya but not sure
aw k gona try again
let f(t)=t(1-cos(t)) and g(t)=t-sin(t) ----Recall this above--- f(t)=t-tcos(t) f'(t)=(t-tcos(t))'=(t)'-(tcos(t))' Derivative of t is easy For finding the derivative of tcos(t) you need the product rule. g'(t) is a bit easier to find.
ZZZ dont get this keep on getting 0/0
You should not getting 0/0 after differentiating three times.