So I have worked out this forumla
$\displaystyle N = \frac{1}{3sin^2x} * (1 - cos^3x)$
As x tends to zero, N tends to 0.5, however if x =0, N = 0. Is there anyway to prove that N should be 0.5 when x = 0?
Thanks
Calypso
Well, strictly speaking, you can't "prove that N should be 0.5". It's not, nor is N= 0- it's undefined at x= 0. It's limit is 0.5, as x goes to 0, yes. The function, as you write it, is not continuous at x= 0. If you want to make that function continuous at x= 0, you would have to redefine it to be 0.5 at x= 0.
Again, Tonio beat me by 1 minute!