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)

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- Sep 23rd 2009, 08:48 PMkinproblem 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 PMmr fantastic
- Sep 24th 2009, 12:36 AMkin
- Sep 24th 2009, 02:05 AMmr fantastic
- Sep 24th 2009, 02:46 AMkin
- Sep 24th 2009, 02:48 AMmr fantastic