Limit of Function of Convergent Sequence Let $a_n$ be a real valued sequence.  Assume that 
$a_n\to0$ as $n\to\infty$.  Use this to prove that $(1+a_n/n)^n\to1$ as $n\to\infty$.
This proof would be simple if it wasn't for that pesky exponent.  I suspect that I"m supposed to use the Bernoulli inequality to solve this somehow, along with monotone convergence theorem.  But I'm just really struggling to make any progress with this.  Any advice?
 A: Let observe that the limit trivially holds for $a_n=0$, thus without loss of generality assume $a_n\neq 0$.
1. Proof by log
Note that
$$(1+a_n/n)^n=e^{n\log(1+a_n/n)}\sim e^{a_n}\to 1$$
indeed 
$$\log(1+a_n/n)=\frac{a_n}{n}\frac{\log(1+a_n/n)}{a_n/n}\sim \frac{a_n}{n}$$
2. Proof by squeeze theorem
As an alternative note that by Bernoulli and by $\left(1+\frac{1}{n}\right)^{n}\le e$ we have that
$$1+a_n=1+n\frac{a_n}{n}\le \left(1+\frac{a_n}{n}\right)^n\le \left(1+\frac{|a_n|}{n}\right)^n=\left[\left(1+\frac{|a_n|}{n}\right)^{\frac{n}{|a_n|}}\right]^{|a_n|}\le e^{|a_n|}$$
thus by squeeze theorem 
$$\left(1+\frac{a_n}{n}\right)^n\to 1$$
A: (i). For $x\geq 0$ we have $$0\leq \ln (1+x)=\int_1^{1+x}\frac {1}{y}dy\leq \int_1^{1+x} 1dy=x.$$ (ii).  For $z\in (-1,0)$ we have $$0> \ln (1-z)=\int_1^{1-z}\frac {1}{y}dy>\int_1^{1-z}\frac {1}{1-z}dz= \frac {-z}{1-z}.$$ 
(iii). Since $a_n\to 0$ we have  $|a_n/n|<1$ for all but finitely many $n.$ 
(iv). If $a_n\geq 0$ then by (i) we have $$0\leq \ln ((1+a_n/n)^n)=n\ln (1+a_n/n)\leq  n(a_n/n)=a_n=|a_n|.$$ (v).  If $-1<a_n/n$ then  by (ii) we have $$0>\ln ((1+a_n/n)^n)=n\ln (1+a_n/n)=n\ln (1-|a_n/n|)>n\frac {-|a_n/n|}{1-|a_n/n|}=$$ $$=-\frac {|a_n|}{1-|a_n|/n}.$$ (vi). By (iv) and (v), when  $|a_n/n|<1$ we  have $$|\ln ((1+a_n/n)^n)|\leq \max\left(|a_n|, \frac  {|a_n|}{1-|a_n|/n}\right)=\frac {|a_n|}{1-|a_n|/n}.$$  Therefore $\lim_{n\to \infty} \ln ((1+a_n/n)^n)=0.$
