t - 0 (t(1-cos(t))/(t -sin(t))
use using the lhoptialrule
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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 )
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@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.
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
----Recall this above---
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.