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.


Herein, we present an approach that for any given $\epsilon>0$, produces a number $N$, which depends on $\epsilon$ and not $x$, such that $\displaystyle \left|e^x-\left(1+\frac xn\right)^n\right|<\epsilon$ whenever $n>N$.


To do this we will use the inequalities, which I established in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality.  The inequalities used in the ensuing analysis are


$$\bbox[5px,border:2px solid #C0A000]{e^x\le 1+x} \tag 1$$


for $x<-1$ and


$$\bbox[5px,border:2px solid #C0A000]{\log(x)\ge \frac{x-1}{x}} \tag 2$$


for $x>0$.


To this end, we proceed.


We assume that $x\in [a,b]$ and that $\epsilon>0$ is given. Furthermore, we will choose $n$ such that $n>-x$ for all $x\in[a,b]$.
Using $(1)$ and $(2)$ we can write
$$\begin{align}
\left|e^x-\left(1+\frac xn\right)^n\right|&=\left|e^x-e^{n\log\left(1+\frac xn\right)}\right|\\\\
&\le \left|e^x-e^{\frac{x}{1+x/n}}\right|\\\\
&=e^x\,\left|1-e^{-x^2/(x+n)}\right|\\\\
&\le e^x\,\left|\frac{x^2}{x+n}\right|\\\\
&\le e^{b}\frac{|\max^2(a,b)|}{n+a}\\\\
&<\epsilon
\end{align}$$
whenever $ \displaystyle n>\frac{e^b\,\max^2(a,b)}{\epsilon}-a$.  We take $\displaystyle N(\epsilon)=1+\left\lfloor \frac{e^b\,\max^2(a,b)}{\epsilon}-a \right\rfloor$ and we are done!
