Prove that:
$\displaystyle \frac{(1+\frac{1}{n})^x}{1+\frac{x}{n}}=1+\frac{x( x-1)}{2n^2}+o\left(\frac{1}{n^2}\right)$
I believe that $\displaystyle (1+y)^\alpha=1+\alpha y+\alpha(\alpha-1)y^2/2+o(y^2)$ (perhaps the $\displaystyle o(y^2)$ part needs to be double-checked). Expanding the nominator and the denominator in this way gives you the right-hand side.
You are interested in the rate of growth of the left hand side as $\displaystyle n \to \infty$ so put $\displaystyle \varepsilon=1/n$ and expand as a Mclaurin series with the remainder term after the term in $\displaystyle \varepsilon^2$. Changing back to $\displaystyle n$ will give the result.
CB