# problem of limit

• Sep 23rd 2009, 08:48 PM
kin
problem of limit
i try L'Hopital at the first glance,

but i make it more complicated....

$\displaystyle \lim_{n\to\infty}$(2x -sin 2x)/(x (sinx)^2)
• Sep 23rd 2009, 09:26 PM
mr fantastic
Quote:

Originally Posted by kin
i try L'Hopital at the first glance,

but i make it more complicated....

$\displaystyle \lim_{n\to\infty}$(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.
• Sep 24th 2009, 12:36 AM
kin
Quote:

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...........
• Sep 24th 2009, 02:05 AM
mr fantastic
Quote:

Originally Posted by kin
it doesnt work...........

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

Are you sure it's $\displaystyle x \rightarrow + \infty$? (By the way, you have $\displaystyle {\color{red}n} \rightarrow + \infty$ .... ) Because I doubt the limit exists if that's the case.
• Sep 24th 2009, 02:46 AM
kin
Quote:

Originally Posted by mr fantastic
Sorry, I misread the limit as x --> 0.

Are you sure it's $\displaystyle x \rightarrow + \infty$? (By the way, you have $\displaystyle {\color{red}n} \rightarrow + \infty$ .... ) 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..
• Sep 24th 2009, 02:48 AM
mr fantastic
Quote:

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.

..