Doubt in Calculus Limit Problem Determine $\displaystyle{\lim_{n\to\infty} x_n}$ if $$\left(1+\frac{1}{n}\right)^{n+x_n}=e,\forall n\in \mathbb{N} $$
I have typed 2 methods giving two different answers

Method 1
$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n+x_n}=\lim_{n\to\infty}e\\\implies
\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\cdot\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{x_n}=e\\\implies
e\cdot\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{x_n}=e
\\\implies
\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{x_n}=1$$
We know:
$$\lim_{x\to a}\left(f\left(x\right)^{g\left(x\right)}\right)=L\Rightarrow\ln L=g\left(x\right)\lim_{x\to a}\ln\left(f\left(x\right)\right)$$
$$\therefore 0=x_n\lim_{n\to\infty}(\ln(1+\frac{1}{n}))$$
Now $\lim_{n\to\infty}(\ln(1+\frac{1}{n}))=0$ therefore $x_n$ can be anything.

Method 2
Take log both sides and get
$$\left(x_n+n\right)\ln\left(1+\frac{1}{n}\right)=1$$
Also as $$t=\frac{1}{n};n\to\infty;t\to0$$
Now $$x_n=\frac{1}{\ln\left(1+t\right)}-\frac{1}{t}=\frac{t-\ln\left(1+t\right)}{t^2\left(\frac{\ln\left(1+t\right)}{t}\right)}$$
$$\lim_{n\to\infty}x_n=\lim_{t\to0}\frac{t-\ln\left(1+t\right)}{t^2\left(\frac{\ln\left(1+t\right)}{t}\right)}=\lim_{t\to0}\frac{t-\ln\left(1+t\right)}{t^2}\cdot\lim_{t\to0}\frac{1}{\frac{\ln\left(1+t\right)}{t}}=\frac{1}{2}\cdot1$$
(Used certain standard limits which you can solve by series or wikipedia for more methods.)

Please help.
(By the way I typed the note in Mathquill. Really nice to use)
 A: We have $$ \left(1+\frac{1}{n}\right)^{x_n} =e\left(1+\frac{1}{n}\right)^{-n}$$ hence $$x_n =\frac{1}{\ln\left(1+\frac{1}{n}\right)} -n$$ and therefore $$\lim_{n\to\infty} x_n =\frac{1}{2}.$$
A: You are still solving two different problems. In the first case, you are saying:

If I know that $$\lim_{n\to\infty} \left(1+\frac1n\right)^{n+x_n} = e$$ can I determine $\lim_{n\to\infty} x_n$?

Your conclusion is correct, you cannot determine $\lim x_n$ based in this information. For example, if $\forall n x_n=0$, then $\lim_{n\to\infty} \left(1+\frac1n\right)^{n+x_n} = e$, and $\lim_{n\to\infty}x_n = 0$, while if $\forall n x_n=1$, it still satisfies the condition.
But the second approach says:

If I know that $$\left(1+\frac1n\right)^{n+x_n} = e$$ can I determine $\lim_{n\to\infty} x_n$?

This is a lot more information. This entirely specifies the values $x_n$, while the previous condition only gives you some really loose information about the sequence. So it is no surprise that you can conclude more about $x_n$ if you have the entire sequence rather than just a property that can match lots of different sequences.
A: Issues
In your first method, you wrote $\lim_{n\to\infty}(1+\frac{1}{n})^{x_n}$ without proving its existence, so that line is already incorrect, even though it turns out to be still true. In a later line you wrote an expression where $x$ occurred outside a limit that binds $x$. That is just nonsense because you cannot talk about the value of a variable outside its scope.
In your second method, you got the correct answer, but the method is still not quite right. You did not prove that you can expand the domain on which you take limits to a punctured neighbourhood of $0$ (small non-zero reals). Originally the limit was for integer $n$, which would correspond to reciprocal values of $t$, which certainly do not cover all small non-zero reals. So if you obtain a limit of your transformed expression, like here, the answer would be correct because you did cover all reciprocals, but if you do not obtain a limit, it does not mean that the original sequence also does not have a limit.
Asymptotic expansion
As $n \to \infty$:
  $x_n = \dfrac{\ln(e)}{\ln(1+\frac{1}{n})} - n \in \dfrac{1}{\frac{1}{n}-\frac{1}{2n^2}+Θ(\frac{1}{n^3})}-n$ because $\frac{1}{n} \in o(1)$
  $ = \dfrac{n}{1-(\frac{1}{2n}+Θ(\frac{1}{n^2}))} ⊆ n\bigg(1+(\frac{1}{2n}+Θ(\frac{1}{n^2}))+Θ\left((\frac{1}{2n}+Θ(\frac{1}{n^2}))^2\right)\bigg)-n$ because:
    $\frac{1}{2n}+Θ(\frac{1}{n^2}) ⊆ o(1)$
  $ ⊆ n(1+\frac{1}{2n}+Θ(\frac{1}{n^2}))-n = \frac{1}{2}+Θ(\frac{1}{n})$
I encourage you to learn asymptotic expansion because it works on many limits with no ingenuity or tricks necessary.
