Another solution:

By using a trig identity,

By evaluating all limits,

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- June 10th 2008, 12:56 PMwingless
- June 10th 2008, 03:01 PMalbi
- June 10th 2008, 03:39 PMalbi
Try this one (without using Stirling formula and L'Hospital):

- June 10th 2008, 03:56 PMMathstud28
- June 10th 2008, 04:30 PMgalactus
Take a look at

- June 10th 2008, 05:01 PMMathstud28
- June 10th 2008, 06:17 PMMathstud28
Try this one without L'hopitals

- June 10th 2008, 08:18 PMMathstud28
- June 10th 2008, 10:08 PMMathstud28
- June 11th 2008, 09:03 AMalbi
- June 11th 2008, 09:28 AMPaulRS
- June 11th 2008, 11:38 AMMathstud28
Yeah, I realized this, and I was unsure when I posted it, but I thought "Hey if I am wrong someone will correct me" and here you did. So what would your solution be?

I am actually kind of stumped without L'hopital's. If it is something easy I am going to kick myself

Actually

So then we have that

If

Now letting

and as

So we have

Which by the root/ratio test trick

we get

this limit is equal to

And since it can be seen that as that

and it can also be said that

So we have that this whole mess above is equivalent to

So now we have

(Sun)

2(Sun)(Sun)(Sun)th post! - June 11th 2008, 12:32 PMMathstud28
- June 11th 2008, 02:40 PMMathstud28
Here is another easy one

- June 11th 2008, 03:42 PMKrizalid
Try this one: