What is the Limit of (1+x)^(1/x) as x approaches 0?

[vkash]

Homework Statement

see attachment

Homework Equations

lim_x->0(1+x)^1/x=e

The Attempt at a Solution

step 1: take log in both sides of a_n=...(as in function).
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 lim_n->infln a_n 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

 

[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.

 

[vkash]

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.

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

 

[Dick]

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

Of course, sorry.

 

[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.

 

[vkash]

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.

doesn't understand>

 

[Dick]

doesn't understand>

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.