Thread: problem of limit

i try L'Hopital at the first glance,

but i make it more complicated....

(2x -sin 2x)/(x (sinx)^2)

Originally Posted by kin

i try L'Hopital at the first glance,

but i make it more complicated....

(2x -sin 2x)/(x (sinx)^2)

It probably requires three applications of l'Hopital's Rule. Personally, I'd substitute the power series for sin(x) and sin(2x), simplify and then take the limit.

Originally Posted by mr fantastic

It probably requires three applications of l'Hopital's Rule. Personally, I'd substitute the power series for sin(x) and sin(2x), simplify and then take the limit.

it doesnt work...........

Originally Posted by kin

it doesnt work...........

Sorry, I misread the limit as x --> 0.

Are you sure it's ? (By the way, you have .... ) Because I doubt the limit exists if that's the case.

Originally Posted by mr fantastic

Sorry, I misread the limit as x --> 0.

Are you sure it's ? (By the way, you have .... ) Because I doubt the limit exists if that's the case.

sorry, it should be x -->infinity

there are two cases:

1) x tends to 0: my answer = 4/3 ( am i correct?)

2) x tends to infinity: i hv no idea..

Originally Posted by kin

there are two cases:

1)limit tends to 0: my answer = 4/3 ( am i correct?) Mr F says: Yes.

2)limit tends to infinity: i hv no idea.. Mr F says: I doubt the limit exists.

..