An interesting limit

on a hunch, is it 1/2?

indeed it is:) justification?

Ok, i worked with this reckoning that as x becomes large, each of the factors are almost equal so that \sqrt [n] {(x+2)(x+4)...(x+2n)} will tend to the AM of the factors which is x+n+1.

Likewise, the second term tends to x+n+1/2

Hence their difference is 1/2

Now, between the earlier post and this I arrived at the proof of

\sqrt [n] {(x+a_1)(x+a_2)...(x+a_n)} \rightarrow x + A as x \rightarrow \infty where

A = \frac{a_1+a_2+...+a_n}{n}

This can easily be attempted by the students. So, letz wait

Yes... you are right (which you usually are [1]). And of course, if one proves that last result of yours, he/she will reach the conclusion. Lets wait.

{(x+a₁)(x+a₂)...(x+a_n)}^1/n = x+A, as x →∞, if the proof of this is reqd, it's damn easy....
Consider f(x)=R.H.SL.H.S, now dividing N & D by x we have our result,

But is this the Qsn at all...or is to find the limit as x→∞ of {(x+a₁)(x+a₂)...(x+a_n)}^1/n ???

here is my sir's solution (you can download the file)
http://www.mediafire.com/?ge3zu7zt2yh