1. ## evaluating intergral

So I have worked out this forumla

$N = \frac{1}{3sin^2x} * (1 - cos^3x)$

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

2. Originally Posted by calypso
So I have worked out this forumla

$N = \frac{1}{3sin^2x} * (1 - cos^3x)$

As x tends to zero, N tends to 0.5, however if x =0, N = 0.

Says who? N isn't defined at x = 0 because you'd get $\frac{0}{0}$ , so you can define N(0):= 0.5 .

Tonio

Is there anyway to prove that N should be 0.5 when x = 0?

Thanks

Calypso
.

3. Originally Posted by calypso
So I have worked out this forumla

$N = \frac{1}{3sin^2x} * (1 - cos^3x)$

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!

4. Is it possible to use L'hopital rule to find answer at x = 0?

5. Originally Posted by calypso
Is it possible to use L'hopital rule to find answer at x = 0?

Yes. That's how I did it.

Tonio

6. thanks