Let $\{a_{n}\}$ be any sequence of reals such that $\lim \limits_{n\rightarrow \infty} na_{n} =0$... Let $\{a_{n}\}$ be any sequence of reals such that $\lim \limits_{n\rightarrow  \infty } na_{n} =0$. Prove that $$\lim_{n \rightarrow \infty } \left( 1 + \frac{1}{n} + a_{n}\right)^{n} =e$$
I think I have shown that  $\lim \limits_{ n\rightarrow \infty} \{ a_{n} \}$ must be zero 
and. I am thinking maybe showing the  sequence above is bounded and decreasing might help. Basically I want to show it has the same limit as $\lim  \limits_{n \rightarrow \infty } ( 1 + \frac{1}{n}) $ which is defined as e in my book.
 A: One of the standard arguments goes as follows: Fix a sufficiently small $\varepsilon > 0$. Then by the assumption, there exists $N$ such that
$$ -\frac{\varepsilon}{n} \leq a_n \leq \frac{\varepsilon}{n} $$
for all $n \geq N$. Subsequently, we have
$$ \left(1 + \frac{1-\varepsilon}{n}\right)^{n} \leq \left(1 + \frac{1}{n} + a_{n}\right)^{n} \leq \left(1 + \frac{1+\varepsilon}{n}\right)^{n}. $$
In view of the limit
$$ \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^{n} = e^{x}, $$
we have
$$ e^{1-\epsilon} \leq \liminf_{n\to\infty} \left(1 + \frac{1}{n} + a_{n}\right)^{n} \leq \limsup_{n\to\infty} \left(1 + \frac{1}{n} + a_{n}\right)^{n} \leq e^{1+\epsilon}. $$
Since this is true for any small $\varepsilon > 0$. we conclude that
$$ \liminf_{n\to\infty} \left(1 + \frac{1}{n} + a_{n}\right)^{n} = \limsup_{n\to\infty} \left(1 + \frac{1}{n} + a_{n}\right)^{n} = e. $$
Therefore the convergence follows.
A: Another approach:
If $\lim\limits_{x\to{+\infty}} f(x)^{g(x)}$ is $1^{+\infty}$, which is an indeterminate form, then: $$\lim_{x\to{+\infty}} f(x)^{g(x)}=e^{\lim\limits_{x\to +\infty}\big(f(x)-1\big)g(x)}$$
A: First, let's factor out the part we know, then simplify
$$ \begin{align}
\lim_{n \rightarrow \infty } \left( 1 + \frac{1}{n} + a_{n}\right)^{n} 
&= \lim_{n \rightarrow \infty } \left( 1 + \frac{1}{n} \right)^{n} 
\left( \frac{1 + \frac{1}{n} + a_{n}}{1 + \frac{1}{n}} \right)^{n}
\\ &= \lim_{n \rightarrow \infty } \left( 1 + \frac{1}{n} \right)^{n} 
\lim_{n \rightarrow \infty }
\left( \frac{1 + \frac{1}{n} + a_{n}}{1 + \frac{1}{n}} \right)^{n}
\\ &=e \lim_{n \rightarrow \infty } \left(1 +  \frac{n a_{n}}{n + 1} \right)^{n}
\end{align}$$
What's left looks sort of like the usual limit for $e$, so let's fiddle with it
$$ \begin{align}
\cdots &=e \lim_{n \rightarrow \infty } \left(1 +  \frac{(n-1) a_{n-1}}{n} \right)^{n-1}
\\ &= e \lim_{n \rightarrow \infty } \left(\left(1 +  \frac{(n-1) a_{n-1}}{n} \right)^{n} \right)^{\frac{n-1}{n}}
\\ &= e \left( \lim_{n \rightarrow \infty } \left(1 +  \frac{(n-1) a_{n-1}}{n} \right)^{n} \right)^{\lim_{n \rightarrow \infty } \frac{n-1}{n}}
\\ &= e \left( \lim_{n \rightarrow \infty } \left(1 +  \frac{(n-1) a_{n-1}}{n} \right)^{n} \right)^{1}
\\ &= e \lim_{n \rightarrow \infty } \left(1 +  \frac{(n-1) a_{n-1}}{n} \right)^{n} 
\end{align}
$$
Better.  This looks even more like the limit for $e^x$; but the part that should be $x$ is going to $0$.  So let's call it $L$ and bound it:
$$ \begin{align}
L &\leq e \lim_{n \rightarrow \infty } \left(1 +  \frac{x}{n} \right)^{n} 
\\ &= e^{1+x}
\end{align}
$$
$$ \begin{align}
L &\geq e \lim_{n \rightarrow \infty } \left(1 - \frac{x}{n} \right)^{n} 
\\ &= e^{1-x}
\end{align}
$$
for all $x > 0$. That is, if $L$ is the limit, then
$$ e^{1-x} \leq L \leq e^{1+x} $$
Now we can use the method of exhaustion (i.e. squeeze theorem):
$$ e = \lim_{x \to 0^+} e^{1-x}  \leq L \leq \lim_{x \to 0^+} e^{1+x} = e$$
and therefore $L = e$.
A: Hint: Rewrite the expression as 
$$\left(\left(1+\frac{1+na_n}{n}\right)^{\frac{n}{1+na_n}}\right)^{1+na_n}.$$
