# Limit (Taylor expansion?)

1. Jul 20, 2015

### Incand

1. The problem statement, all variables and given/known data
Let $x_n$ be the solution to the equation
$\left( 1+\frac{1}{n} \right)^{n+x} = e$

Calculate $\lim_{n\to \infty} x_n$

2. Relevant equations
N/A

3. The attempt at a solution
Since $\lim_{n \to \infty} \left(1+ \frac{1}{n} \right) = e$ that tells me that $\lim_{n\to \infty} x_n = 0$ but the answer in the book says it should be $\frac{1}{2}$ which I don't understand at all.
This is also in a section about Taylorexpansion which suggest I should use that somehow. If I expand on x i get something like this
$\left( 1+\frac{1}{n} \right)^{n+x} = \left( 1+\frac{1}{n} \right)^{n}\cdot \left( 1+\frac{1}{n} \right)^{x} =\left( 1+\frac{1}{n} \right)^{n}\cdot (1+\ln (1+\frac{1}{n})x + O(x^2) )$
which doesn't seen to help me and I still don't see why $x_n\to \frac{1}{2}$.

Last edited: Jul 20, 2015
2. Jul 20, 2015

### pasmith

What you are given is $$\left(1 + \frac1n\right)^{n+x_n} = e.$$ The first step is to solve for $x_n$ in terms of $n$. Then you can consider how you might determine the limit.

3. Jul 20, 2015

### Incand

Alright I think I got it.
If I take the natural logarithm on both sides i get
$(n+x_n)\left( 1 + \frac{1}{n} \right) = 1$
seperating $x_n$ i get
$x_n = \frac{1-n\ln (1+\frac{1}{n}) }{\ln(1+\frac{1}{n})}$
and If I use taylor expansion $\ln (1+ \frac{1}{n} ) = \frac{1}{n} - \frac{1}{2n^2} + O(\frac{1}{n^3})$ i get (with $n\to \infty$)
$x_n = \frac{1-n(\frac{1}{n}-\frac{1}{2n^2} + O(\frac{1}{n^3})) } {\frac{1}{n} + O(\frac{1}{n^2})} = \frac{\frac{1}{2n}}{\frac{1}{n}} = \frac{1}{2}$.

Thanks! Any idea where I go wrong In my initial argument where I thought that since $\left( 1 + \frac{1}{n} \right)^{n} = e$ should mean $x_n=0$?

4. Jul 20, 2015

### Incand

Edit: Accidently posted twice

5. Jul 20, 2015

### RUber

I think it might be hidden in the word "solution".
Clearly, the limit of $\left( 1 + \frac 1n \right)^{n} = e$, but really, any constant in the exponent would have the same limit.
You have to solve for x in order to determine the pattern.

6. Jul 20, 2015

### Incand

Thank you, that makes sense!