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 
$(1+x^2/n^2)^n -1 = x^kO(x/n).$

$$A_n=\left(1+\frac{x^2}{n^2}\right)^n\implies \log(A_n)=n \log\left(1+\frac{x^2}{n^2}\right)$$ Now, using Taylor series for large values of $n$ $$\log\left(1+\frac{x^2}{n^2}\right)=\frac{x^2}{n^2}-\frac{x^4}{2 n^4}+O\left(\frac{1}{n^6}\right)$$ $$ \log(A_n)=\frac{x^2}{n}-\frac{x^4}{2 n^3}+O\left(\frac{1}{n^6}\right)$$ Taylor again using $$A_n=e^{\log(A_n)}=1+\frac{x^2}{n}+\frac{x^4}{2 n^2}+\frac{x^4 \left(x^2-3\right)}{6 n^3}+\frac{x^6
   \left(x^2-12\right)}{24 n^4}+O\left(\frac{1}{n^5}\right)$$ For illustration purposes, let us use $x=10$ and $n=100$ $$A_{100}=\left(\frac{101}{100}\right)^{100}\approx 2.70481$$ while the above expansion would give $$\frac{1619}{600}\approx 2.69833$$
