Limit of a function involving a sequence. I have the following problem:

Suppose that $\lim_{n \to \infty} a_n = 0$. Prove that for any $x$
  $$\lim_{n \to \infty} \left(1+a_n \frac{x}{n}\right)^n = 1.$$

I have tried replacing $a_n$ with a function $a(n)$ and applying L'Hopital's Rule but got no useful result. I also tried going directly to the definitions and showed that $\lim_{n \to \infty} (1+a_n \frac{x}{n}) = 1$, but I don't know how to deal with the $n$th power (that makes the limit $\lim_{n \to \infty} \exp(n(1+a_n \frac{x}{n}))$ indeterminate).
 A: Take log:
$$
\lim_{n\to\infty} n\ln\left(1 + a_n\frac{x}{n}\right).
$$
But if $a_n\to0$
$$
\ln\left(1 + a_n\frac{x}{n}\right)\sim a_n\frac xn,
$$
and
$$
\lim_{n\to\infty} n\ln\left(1 + a_n\frac{x}{n}\right) = \lim_{n\to\infty} a_n x = x\lim_{n\to\infty} a_n = 0;
$$
so
$$
\lim_{n\to\infty} \left(1 + a_n\frac{x}{n}\right)^n = 1
$$
A: Notice, $$\lim_{n\to \infty}\left(1+a_n\frac{x}{n}\right)^n$$
Using binomial expansion of $\left(1+a_n\frac{x}{n}\right)^n$ & neglecting higher power terms, we get 
$$=\lim_{n\to \infty}\left(1+na_n\frac{x}{n}\right)$$
$$=\lim_{n\to \infty}\left(1+a_n x\right)$$ $$=\lim_{n\to \infty} 1+x\lim_{n\to \infty}a_n $$ $$= 1+x(0)=1 $$
A: I am giving an simple answer.
$\lim\limits_{x\to \infty} a_n \to 0$ 
Now, $\lim\limits_{x \to \infty} (1+a_n \frac xn)^n =\lim\limits_{x\to \infty}((1+a_n \frac xn)^ {\frac n{a_n x}})^{a_nx} =e^{\lim\limits_{n\to \infty}\;a_n x} =e^0=1$
A: $$
\left(1 + \frac{a_n x}{n}\right)^n = 
\bigg[
\underbrace{\left(1+\frac{a_n x}{n}\right)^{\frac{n}{a_n x}}}_{\to e}
\bigg]^{\overbrace{a_n x}^{\to 0}}
\to e^0 = 1
$$
A: Fix $x\in\mathbf R$.
By Bernoulli's inequality (here $n$ has to be large enough, so that $a_n x/n>-1$),
$$
\Bigl(1+\frac{a_nx}{n}\Bigr)^n\geq 1+a_n x. 
$$
On the other hand, since
$$
\Bigl(1+\frac{a}{n}\Bigr)^n\leq e^a
$$
for all $a\in\mathbf R$,
$$
\Bigl(1+\frac{a_nx}{n}\Bigr)^n\leq e^{a_n x}.
$$
Now use the sandwich theorem.
