some quick help for a friend

[steve_254]

could someone do out this limit please?

limx->0 [ sin^3(3x) / sin^3(2x) ]

 

[Guest]

limx->0 [ sin^3(3x) / sin^3(2x) ]

ok here is a start for you....

limx->0 ( u/v) where u=sin^3(3x) and v= sin^3(2x)

working just on the top line need to find the du/dx of this to start with.

u = a^3 where a= sin 3x

da/dx = 3 cos 3x
du/da = 3 a^2

then du/dx = du/da . da/dx

= 3 (sin 3x)^2 . (3 cos 3x)
= 9 (sin ^2(3x) ) (cos 3x )

do the same process for the v =.....

then apply the dy/dx = (vdu/dx - u dv/dx) / v^2

does this help you to start????

 

[steve_254]

any possibility of doin this problem without derivitives ?

 

[Guest]

given you using the concept of lim x>0 I don't know how you can move away from derivities.....sorry.

 

[stapel]

any possibility of doin this problem without derivitives ?

Yes:

. . .sin³(3x) / sin³(2x)

. . . . .= (8/27) (27/8) (sin³(3x) / sin³(2x))

. . . . .= (27/8) (sin³(3x) / 27) (8 / sin³(2x))

. . . . .= (27/8) (sin(3x)/3)³ (2/sin(2x))³

Now use the limit they gave you for sin(ß)/ß as ß goes to zero.

Eliz.

 

[Guest]

very smooth and neat -well done