anonymous
  • anonymous
limit as x tends to infinity of 5sin(x^3)/sin^3(2x)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
sorry, x tends to 0
anonymous
  • anonymous
That makes more sense.
anonymous
  • anonymous
yeah lol..any ideas?

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anonymous
  • 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
  • anonymous
ok, i'll give that a go
anonymous
  • anonymous
done the top and got 15x^2cosx^3... the bottom could take a while longer
anonymous
  • 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
  • anonymous
is there any sin/cos equations for sin^3(x) that can be used
anonymous
  • anonymous
its just gonna be a long retricel'hopital iteration i think
anonymous
  • anonymous
it cant be that difficult, its only worth 2 marks lol
anonymous
  • 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
  • anonymous
well l'hopital's not really that difficult so meh could be worth 2 marks
anonymous
  • anonymous
ive applied l'hopitals once and it diddnt help :(
anonymous
  • anonymous
Keep going, sometimes it takes a few. The problem is probably aimed to teach you that.
anonymous
  • anonymous
think ive got it
anonymous
  • anonymous
sin^3(2x)=(2sinxcosx)^3=8sin^3xcos^3x so we have 5sin(x^3)/(8sin^3xcos^3x)
anonymous
  • anonymous
oh wait, nope :(
anonymous
  • anonymous
answers 5/8 btw
anonymous
  • anonymous
yeah, i got that off wolfram..there must be a way of getting sin(x)/x in there somehow

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