hey guys i checked with my sir, it seems the answer is indeed 1.

 

  It is easy to see that this limit is equal to 1.

Now let  

 

 

=>  

 

Also, by AM-GM:

So we have proved:

Hence, since , we'll get,

by sandwich theorem, that

I am not very sure about the other limit. A binomial expansion suggests the limit is 1. But I dont think there is anything wrong in  my working above either.

layman, do you mind provididng the solution?

i happened to come across this limit somewhere (maybe in one of my parent's books)  . My proffessor told me about sterling's approximation and used it to prove the limit.

 

 

For the 4th question,

 

edited (made a horrible mistake)

 

For the 5th one,

 

By Stirling's formula is approximately equal to , so it looks like  should be approximately . But this looks like a very tricky thing to establish with a rigorous proof..

 

What is the source of the problem?

Let x y z be the number.
Any one of the nine digits from 1 to 9 can be the value of 'x' . But for 'y' , zero can also be a choice. Hence if there was no restriction on the number, 'y' could have assumed 10 different numbers. Now we know that, whatever value was chosen for 'x', the same value cannot be chosen for 'y'. hence only 9 possibilities remain for 'y'. Similarly, 8 different possibilities are there for 'z'.

 

Hence there  9 x 9 x 8 = 648 such numbers.

 

 

How did you get so many values?

nice sir... the above is the method i had in mind.

sandeep, last time he said its not open to everyone..so just guess it might be the same this time too.

Anyway, i wont let this weakness let me down again

 

 

In general,

hence inequality simplifies to
i.e   for x > 1 which is obviosully true.

 

Similar method for x < 1 , with the differnce that .

unfair!

u both didnt even wait for him to say if it was open to all!!