Is my proof that if $x_n$ is a sequence such that $x_n \rightarrow +\infty$ then $(1+\frac{1}{x_n})^{x_n}\rightarrow e$ correct? I'm trying to prove that if $x_n$ is a sequence such that $x_n \rightarrow +\infty$ then $(1+\frac{1}{x_n})^{x_n}\rightarrow e$. Here is what I've done so far:
We know that $\{(1+\frac{1}{n})^n\}_{n \geq 1} \rightarrow e$ so it's true that $\forall_{\epsilon >0}\exists_{n\geq 1}\forall_{n'\geq n}|(1+\frac{1}{n})^n-e|<\epsilon$. We will use that to pick $n'$ for a given $\epsilon$ later.
Now we know that $x_n \rightarrow +\infty$, so it is true that $\forall_{\epsilon > 0}\exists_{n' \geq n_0}\forall_{n\geq n'} x_n \geq \epsilon$.
Let's analyze $(1+\frac{1}{x_n})^{x_n}$ now. We want to show that it tends to $e$ so we're trying to show that
$$\forall_{\epsilon >0}\exists_{n'\geq n_0}\forall_{n\geq n'}|(1+\frac{1}{x_n})^{x_n}-e|<\epsilon$$
So we pick any $\epsilon$. And now we know that for sufficiently large $n$ (let's say $n\geq n'$) $|(1+\frac{1}{n})^n-e| < \epsilon$. And also for any given number we can make $x_n$ bigger than it so let's pick $n''$ such that $\forall_{n\geq n''} x_n > n'$. And that's it - for a given $\epsilon$ we've found $n''$ such that $\forall_{n\geq n''}|(1+\frac{1}{x_n})^{x_n}-e|<\epsilon$, which proves that $(1+\frac{1}{x_n})^{x_n}$ tends to $e$.
So is my solution correct? Also - in the workbook there was a hint that we should use squeeze theorem and the property that if a sequence has a limit then for all its subsequences their limit is the same but I don't see that. Does that lead to a better solution?
 A: Your proof is much complicated. A simple and short argument is desirable.
If all of $x_n $ are integers, there is nothing to prove. We assume that $(x_n)$ is a sequence of real numbers.
Take a sequence $(m_n)$ of integers with $m_n\le x_n< m_n+1$ for $n=1, 2,...$ .
Since $$1+\frac{1}{m_n+1}<1+\frac{1}{x_n}\le 1+\frac{1}{m_n},$$ we have 
$$\left(1+\frac{1}{m_n+1}\right)^{m_n}<\left(1+\frac{1}{x_n}\right)^{x_n}\le \left(1+\frac{1}{m_n}\right)^{m_n+1}.$$
Here $(m_n)$ is a subsequence of $(n)_{n\ge 1}$, and hence we have $\lim_{ n\to \infty} \left(1+\frac{1}{m_n}\right)^{m_n}=e$. 
Thus we have 
$$\lim_{ n\to \infty} \left(1+\frac{1}{m_n+1}\right)^{m_n}=\lim_{ n\to \infty} \frac{\left(1+\frac{1}{m_n+1}\right)^{m_n+1}}{\left(1+\frac{1}{m_n+1}\right)}=e$$ 
and
$$\lim_{ n\to \infty} \left(1+\frac{1}{m_n}\right)^{m_n+1}=\lim_{ n\to \infty} \left(1+\frac{1}{m_n}\right)^{m_n}\left(1+\frac{1}{m_n}\right)=e.$$
Using the squeeze theorem, we have $ \lim_{ n\to \infty} \left(1+\frac{1}{x_n}\right)^{x_n}=e$.
