Prove that $\lim_{n\to \infty} \left(1+a_n(x/n)\right)^n=1$ given that $\lim_{n\to \infty} a_n = 0$. I realize that this question already has an answer elsewhere and now realize what the correct answer is (substituting into the limit form of the exponential function and then using continuity of the exponential function to interchange operators).
However I would like to share my incorrect attempt because I do not know exactly why it is incorrect. In particular there is one step to look at.
First I expanded the power $(1 + a_nx/n)^n$ out into a product and wrote "n times" below that. Then I used the fact that the limit of a product is a product of limits. However this step is not justifiable because how does one perform an infinite expansion?
(After that it's straightforward if one overlooks the above. For this product of limits, for each factor expand the limit of a sum into a sum of limits. So now you have an product of $n$ sums: $\lim 1 + \lim a_nx/n$ n times. Finally you put the summand on the right as a product of limits so that this right summand becomes zero. Then you are left with a product of n limits $\lim 1$ so that the answer is 1.)
 A: You did this:
$$\lim_{n\to \infty}\left(1+\frac{a_nx}{n}\right)^n=\lim_{n\to \infty}\underbrace{\left(1+\frac{a_nx}{n}\right)\left(1+\frac{a_nx}{n}\right)\cdots\left(1+\frac{a_nx}{n}\right)}_{n\text{ times}} $$
and because $\lim \frac{a_nx}{n}=0$ then the limit must be $1^n=1$. the mistake here is the fact that the number of terms of the product depends on $n$, if the terms of product does not depend on $n$ then you can use expansion on product.
In other words you did the following step:
$$\lim_{n\to \infty}x_n^n=\lim_{n\to \infty}\underbrace{ x_nx_n\cdots x_n}_{n \text{ times}}=(\lim_{n\to \infty} x_n)^n$$
but as you know in the last term we have an $n$ (in the exponenet) outside of the limit and $n$ is defined only inside the limit
A: Using the reasoning in the argument above, you should get
$$
\lim_{n\to\infty}\left(1+\frac1n\right)^n=1
$$
because each of the $n$ terms of the product tend to $1$. However, this limit is $e=2.718281828\ldots$ This reasoning fails since $1^\infty$ is an indeterminate form, so you can't assume that just because each term of an infinite product goes to $1$ that the product is $1$.

Here is a proof using the inequality $1+x\le e^x$.
Since $\frac1{1+x}=1-\frac x{1+x}\le e^{-\large\frac x{1+x}}$, we get
$$
e^{\large\frac x{1+x}}\le1+x\le e^x
$$
Therefore,
$$
e^{\frac{a_nx}{1+\frac{a_nx}n}}\le\left(1+\frac{a_n x}n\right)^n\le e^{a_nx}
$$
By the Squeeze Theorem, we have
$$
\lim_{n\to\infty}\left(1+\frac{a_n x}n\right)^n=1
$$
A: Probably the simplest way is to observe that $a_n \frac{x}{n} \to 0$ and hence
$$ \lim_n \left(1+a_n(x/n)\right)^{\frac{n}{xa_n}}=e$$
If you want to lose the logarithms, note that 
$$\lim_n \frac{\ln (1+a_n\frac{x}{n})-\ln1}{a_n\frac{x}{n}}=\ln'(1)=1$$
A: $\lim \space (1+a_n \frac x n)^n = \lim \space [(1+a_n \frac x n)^{\frac n {x a_n}}]^{a_n \frac x n} = \mathbb{e}^{\lim \space a_n \frac x n} = \mathbb{e}^0 =1$
