How can i show that both limits exist and are equal to 0? I attached

[ozlem675]

How can i show that both limits exist and are equal to 0? I attached limits as bmp file

"i" refers to "i times" not complex number so you can ignore. This formula is calculated after analysis of algortihms by using summations.

 

[SammyS]

How can i show that both limits exist and are equal to 0? I attached limits as bmp file

"i" refers to "i times" not complex number so you can ignore. This formula is calculated after analysis of algortihms by using summations.

https://www.physicsforums.com/attachment.php?attachmentid=33439&d=1300925084

Is the \sqrt{\log\ n} an exponent?

.

 

[ozlem675]

Yes, it is an exponent

 

[SammyS]

I think the 2nd one is pretty straight forward with L'Hopital's rule.

2^{\sqrt{\log\,x}}=e^{(\log\,2)\sqrt{\log\,x}}

Added in Edit:

Well, I made a mistake, so it's not so easy.

But...

For the first one, try i = 1. That does work using L'Hopital. Then use induction, assuming that i is a positive integer,

 

[ozlem675]

I thought i can solve this limit by L' Hopital's Rule but after derivatives of both parts. The first one is equal to 0 but i have problem with the second one. Can i rewrite log as n^1/2 but I'm not sure about it since the log itself is square root not the "n" nor "2". The second one is equal to 0 after rewriting log but not the first one. Or should i take derivative not once but until the second one is equal to zero which i haven't tried yet since I'm not sure about rewriting log

 

[SammyS]

I thought i can solve this limit by L' Hopital's Rule but after derivatives of both parts.
...
Can i rewrite log as n^1/2 but I'm not sure about it since the log itself is square root not the "n" nor "2"?
...

No.

\frac{1}{2}\,\log(x)=\log(\sqrt{x})

but: \sqrt{\log(x)}\neq\sqrt{x}