Question

lim
t - 0 (t(1-cos(t))/(t -sin(t))
use using the lhoptialrule

Answer 1

*** 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)

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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 )

Answer 2

were did 3x come from

Answer 3

and 2x??

Answer 4

@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

Answer 5

Are you having trouble finding the derivative? That is the only scenario I can see here since it tells you to use lhospital.

Answer 6

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

Answer 7

is the answer 1?

Answer 8

Not for your problem you asked about.

Answer 9

yeah the derivative confused me @myininaya but not sure

Answer 10

aw k gona try again

Answer 11

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.

Answer 12

ZZZ dont get this keep on getting 0/0

Answer 13

You should not getting 0/0 after differentiating three times.