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- anonymous

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

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- anonymous

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

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- anonymous

sorry, x tends to 0

- anonymous

That makes more sense.

- anonymous

yeah lol..any ideas?

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- anonymous

You'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

ok, i'll give that a go

- anonymous

done the top and got 15x^2cosx^3...
the bottom could take a while longer

- anonymous

If 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

is there any sin/cos equations for sin^3(x) that can be used

- anonymous

its just gonna be a long retricel'hopital iteration i think

- anonymous

it cant be that difficult, its only worth 2 marks lol

- anonymous

sin^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

well l'hopital's not really that difficult so meh could be worth 2 marks

- anonymous

ive applied l'hopitals once and it diddnt help :(

- anonymous

Keep going, sometimes it takes a few. The problem is probably aimed to teach you that.

- anonymous

think ive got it

- anonymous

sin^3(2x)=(2sinxcosx)^3=8sin^3xcos^3x
so we have 5sin(x^3)/(8sin^3xcos^3x)

- anonymous

oh wait, nope :(

- anonymous

answers 5/8 btw

- anonymous

yeah, i got that off wolfram..there must be a way of getting sin(x)/x in there somehow

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