What is the limit of this sequence? How does it relate to the exponential function? 
Find $$\lim_{n\to\infty} \left(1+\frac{x^2}{n^2}\right)^n$$

I particular, I am hoping to find the above to be $1+x^kg(x/n)$ where $g(\cdot)$ is a function with uniformly bounded derivatives.
Edit:
I'm trying to prove a result that would be true if 
$(1+x^2/n^2)^n -1 = x^kO(x/n).$

The limit can be evaluated without appeal to the exponential function. To proceed we first note that 
$$\left(1+\frac{x^2}{n^2}\right)^n\left(1-\frac{x^2}{n^2}\right)^n\le 1\tag1$$
Then, rearranging $(1)$, we obtain 
$$ \left(1+\frac{x^2}{n^2}\right)^n\le \frac{1}{\left(1-\frac{x^2}{n^2}\right)^n}\tag 2$$
Using Bernoulli's Inequality on the right-hand side of $(2)$ reveals for $n>|x|$
$$1\le \left(1+\frac{x^2}{n^2}\right)^n\le\frac{1}{1-\frac{x^2}{n}}\tag3$$
whereupon applying the squeeze theorem to $(3)$ yields the coveted limit
$$\lim_{n\to \infty}\left(1+\frac{x^2}{n^2}\right)^n=1$$

Now, we can relate this to the exponential function as follows.  First, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality, that the exponential function satisfies the  inequality
$$ \left(1+\frac xn\right)^n\le e^x \le \frac{1}{1-x} \tag 4$$
for $-n<x<1$.
From $(4)$ we see that
$$\begin{align}
\left(1+\frac{x^2}{n^2}\right)^n&=\left(\left(1+\frac{x^2}{n^2}\right)^{n^2}\right)^{1/n}\\\\
&\le e^{x^2/n}\\\\
&\le \frac{1}{1-\frac {x^2}n}\tag 5
\end{align}$$

Note that $$\frac{1}{1-\frac {x^2}n}=1+\frac{x^2}{n}+O\left(\frac{x^4}{n^2}\right)$$

