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anonymous
 5 years ago
limit as x tends to infinity of 5sin(x^3)/sin^3(2x)?
anonymous
 5 years ago
limit as x tends to infinity of 5sin(x^3)/sin^3(2x)?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That makes more sense.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're going to have to use L'Hopitals rule because if you put 0 in for the Xs you get the indeterminant form 0/0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, i'll give that a go

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0done the top and got 15x^2cosx^3... the bottom could take a while longer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you can somehow determine that the bottom cannot equal zero, you could save yousrself a good deal of time, since the top will always equal zero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is there any sin/cos equations for sin^3(x) that can be used

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its just gonna be a long retricel'hopital iteration i think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it cant be that difficult, its only worth 2 marks lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sin^3 is tricky in the way that you need to use two of the derivation methods you've been taught. Rewrite it as (sin(x)) ^ 3 Now use the power rule + chain rule 3(sin^2(x))(cos x) Another iteration will be required.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well l'hopital's not really that difficult so meh could be worth 2 marks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ive applied l'hopitals once and it diddnt help :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Keep going, sometimes it takes a few. The problem is probably aimed to teach you that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sin^3(2x)=(2sinxcosx)^3=8sin^3xcos^3x so we have 5sin(x^3)/(8sin^3xcos^3x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, i got that off wolfram..there must be a way of getting sin(x)/x in there somehow
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