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.

To find your sequence $(a_n)$ where $|(1 + x/n)^n - e^x| \leqslant a_n \to 0$ -- proving uniform convergence on any bounded interval -- use the inequality $\ln(1+y) \leqslant y$.
We have for $0 \leqslant y < 1$,
$$1+y \leqslant e^y = \sum_{k=0}^{\infty} \frac{y^k}{k!} \leqslant  \sum_{k=0}^{\infty} y^k = \frac1{1-y},$$
Take $y = x/n$. It follows that for $n$ sufficiently large
$$1 + \frac{x}{n} \leqslant e^{x/n} \leqslant \left(1 - \frac{x}{n}\right)^{-1},$$
and
$$\left(1 + \frac{x}{n}\right)^n \leqslant e^x \leqslant \left(1 - \frac{x}{n}\right)^{-n}.$$
The second inequality implies that
$$e^{-x} \geqslant \left(1 - \frac{x}{n}\right)^{n}.$$
Using Bernoulli's inequality $(1 - x^2/n^2)^n \geqslant 1 - x^2/n.$
Hence,
$$0 \leqslant e^{x} - \left(1+ \frac{x}{n}\right)^n = e^{x}\left[1 - e^{-x}\left(1+ \frac{x}{n}\right)^{n}\right]\\ \leqslant e^{x}\left[1 - \left(1+ \frac{x}{n}\right)^{n}\left(1- \frac{x}{n}\right)^{n}\right]\\= e^{x}\left[1 - \left(1- \frac{x^2}{n^2}\right)^{n}\right]\leqslant e^{x}\frac{x^2}{n}.$$
Therefore, for all $x \in [0,K]$, we have as $n \to \infty$  
$$0 \leqslant \left|e^{x} - \left(1+ \frac{x}{n}\right)^n\right| \leqslant e^K\frac{K^2}{n} \rightarrow 0.$$
An almost identical argument for $y \geqslant 0$ shows that
$$0 \leqslant e^{-y} - \left(1- \frac{y}{n}\right)^n \leqslant  e^{-y}\frac{y^2}{n}.$$
Thus if $-y = x \in [-L,0]$ we have
$$0 \leqslant \left|e^{x} - \left(1+ \frac{x}{n}\right)^n\right| = \left|e^{-y} - \left(1- \frac{y}{n}\right)^n\right| \leqslant \frac{L^2}{n} \rightarrow 0.$$
proving uniform convergence on any bounded interval.
