Show that $(1+\frac{x}{n})^n \rightarrow e^x$ uniformly on any bounded interval of the real line. 
Show that $(1+\frac{x}{n})^n \rightarrow e^x$ uniformly on any bounded interval of the real line.

I am trying to argue from the definition of uniform convergence for a sequence of real-valued functions, but am struggling quite a lot. My efforts so far have concentrated on trying to find a sequences,
${a_n}$ which tends to zero, such that 
$$|(1+\frac{x}{n})^n -e^x |\leq a_n$$ for all $n$. But I have been unsuccessful thus far. All help is greatly appreciated.

If you're not familiar with Dini's theorem (I wasn't), work from the power series of $\exp$.  Observe that
$$(1+x/n)^n = \sum_{k=0}^n {n \choose k} \frac{x^k}{n^k}$$
and that
$${n \choose k} \frac{1}{n^k} = \frac{n!}{n^k(n-k)!k!} \to \frac{1}{k!}$$
This means that each coefficient in the expansion of $g_n(x) = \exp(x) - \left(1+\frac{x}{n}\right)^n$ goes to $0$ as $n$ goes to infinity.  Also, all of the coefficients are positive and less than $\frac{1}{k!}$.  
Now, consider the interval $[-R,R]$.  Choose $k> 0$ such that $\sum_{s=k+1}^\infty \frac{R^s}{s!} < \frac{\epsilon}{2}$ (why can you do this?)  For $n$ sufficiently large, we can make each of the first $k+1$ coefficients smaller than $\frac{\epsilon}{2(k+1)(1+R^k)}$, so for any $x \in [-R,R]$,
$$|g_n(x)| \le  \sum_{s=0}^k \frac{\epsilon R^s}{2(k+1)(1+R^k)} + \sum_{s=k+1}^\infty \frac{R^s}{s!} \le \frac{\epsilon}{2} + \frac{\epsilon}{2}.$$
This same reasoning extends (with minor changes) to any bounded interval $[a,b]$ of $\mathbb{R}$ (and even half-open or open intervals).
