At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
sorry, x tends to 0
That makes more sense.
yeah lol..any ideas?
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.
ok, i'll give that a go
done the top and got 15x^2cosx^3... the bottom could take a while longer
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.
is there any sin/cos equations for sin^3(x) that can be used
its just gonna be a long retricel'hopital iteration i think
it cant be that difficult, its only worth 2 marks lol
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.
well l'hopital's not really that difficult so meh could be worth 2 marks
ive applied l'hopitals once and it diddnt help :(
Keep going, sometimes it takes a few. The problem is probably aimed to teach you that.
think ive got it
sin^3(2x)=(2sinxcosx)^3=8sin^3xcos^3x so we have 5sin(x^3)/(8sin^3xcos^3x)
oh wait, nope :(
answers 5/8 btw
yeah, i got that off wolfram..there must be a way of getting sin(x)/x in there somehow