# Solve this limits question

1. Feb 2, 2012

### vkash

1. The problem statement, all variables and given/known data

see attachment

2. Relevant equations

limx->0(1+x)1/x=e

3. The attempt at a solution
step 2: multiply and divide it by 2n inside the log.
step 3: there are n terms in multiplication so substitute 2(of denominator ) with all of the terms.
on solving it comes out to be ln2. which is equal to limn->infln an so final answer will 2.
I solve all the options but none of them give 2.
can you please tell me where am i doing it wrong.
OR
how to know it's answer on wolframalpha. Since i don't know proper commands to do it

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Last edited: Feb 2, 2012
2. Feb 2, 2012

### Dick

The log is (1/x)*log(1+x). I would just use l'Hopital's rule from there. I don't see where step 2 is going.

3. Feb 2, 2012

### vkash

you are solving relevant equation not the question i raised.
see attachment.

4. Feb 2, 2012

### Dick

Of course, sorry.

5. Feb 2, 2012

### Dick

Ok, so taking the log gives you log(1/n)+(1/n)*(log(2n+1)+...+log(2n+n)), ok so far? Now log(1/n)=(1/n)*n*log(1/n)=(1/n)*(log(1/n)+log(1/n)+...+log(1/n)). Add the two sums together and interpret it as a Riemann sum. Sorry to not be more accurate in notation here, but this is along the lines of a hint.

6. Feb 3, 2012

### vkash

doesn't understand>

7. Feb 3, 2012

### Dick

Well, I don't understand how you got ln2 either. If you can show that in more detail maybe someone can tell you why it's wrong. I'm suggesting to write out the sum you get from the log in such a way that it looks like a Riemann sum for an integral.