Question

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

Answer 1

sorry, x tends to 0

Answer 2

That makes more sense.

Answer 3

yeah lol..any ideas?

Answer 4

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.

Answer 5

ok, i'll give that a go

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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.

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

think ive got it

Answer 16

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

Answer 17

oh wait, nope :(

Answer 18

answers 5/8 btw

Answer 19

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