Show that $ \left( 1 + \frac{x}{n}\right)^n$ is uniformly convergent on $S=[0,1]$. 
Show that $ \left( 1 + \frac{x}{n}\right)^n$ is uniformly convergent on $S=[0,1]$.

Given $f_n(x)=\left( 1 + \frac{x}{n}\right)^n$ is a sequence of bounded function on $[0,1]$ and $f:S \rightarrow \mathbb R$ a bounded function , then $f_n(x)$ converges uniformly  to $f$ iff
$$\lim\limits_{n \rightarrow \infty} ||f_n - f|| = \lim\limits_{n \rightarrow \infty}\left(\sup |f_n(x) - f(x)| \right)= 0$$
As $$f(x) = \lim\limits_{n \rightarrow \infty} \left( 1 + \frac{x}{n}\right)^n = e^x$$
We have 
$$\begin{split}
\lim\limits_{n \rightarrow \infty} \left\|\left( 1 + \frac{x}{n} \right)^n - e^x\right\| &= \lim\limits_{n \rightarrow \infty} \left|\sup_{x \in S}\left[\left( 1 + \frac{x}{n}\right)^n - e^x \right ]\right|\\
&= \lim\limits_{n \rightarrow \infty} \left|\left( 1 + \frac{1}{n} \right)^n - e^1 \right|\\
&= \lim\limits_{n \rightarrow \infty} |e-e| \\
&= 0
\end{split}
$$
I am new to sequence. Is this appropriate to show convergence?
Going back to the definition, How can I show that:


*

*"$f_n(x)=\left( 1 + \frac{x}{n}\right)^n$ is a sequence of bounded function on $[0,1]$"?

*$f:S \rightarrow \mathbb R$ a bounded function?
 A: If you don't want to verify the precise maximum you can bound as
$$0 \leqslant e^x - \left(1 + \frac{x}{n} \right)^n =e^x \left[1 -  \left(1 + \frac{x}{n} \right)^ne^{-x}\right] \\ \leqslant  e^x \left[1 -  \left(1 + \frac{x}{n} \right)^n \left(1 - \frac{x}{n} \right)^n\right] \\ =  e^x \left[1 -  \left(1 - \frac{x^2}{n^2} \right)^n \right],$$
since $e^{x/n} > 1 + x/n$ which implies $e^{x} > (1 + x/n)^n $ for $x \in (-n,\infty).$
By Bernoulli's inequality, $(1 - x^2/n^2)^n \geqslant 1 - x^2/n$ and
$$0 \leqslant e^x - \left(1 + \frac{x}{n} \right)^n \leqslant \frac{e^xx^2}{n} \leqslant \frac{e}{n},$$
which enables you to prove uniform convergence on $[0,1]$.
A: Let's write explicitly  the difference 
$$e^x - ( 1 + \frac{x}{n})^n = \sum_{ k \ge 0} \frac{x^k}{k!} - \sum_{k = 0}^n \binom{n}{k}\left(\frac{x}{n}\right )^k=\\
=\sum_{k\ge 0} \frac{[1-\prod_{l=1}^{k-1}(1- l/n)] x^k}{k!}$$.  Since every coefficient $1- \prod_{l=1}^{k-1} (1-\frac{l}{n})$ is $\ge 0$ ( and $0$ for $k \ge n+1$ ) we conclude that 
$$0 < e^x - (1+ \frac{x}{n})^n\le e^1 - (1 + \frac{1}{n})^n$$ for all $n \ge 0$ and $x\in [0,1]$.
Now, it's only necessary to check ( or use ) that $(1+\frac{1}{n})^n \to e$. 
A: As shown in inequality $(2)$ of this answer, $\left(1+\frac xn\right)^n$ is increasing in $n$. Thus, for $x\ge0$,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\left(e^x-\left(1+\frac xn\right)^n\right)
&=e^x-\left(1+\frac xn\right)^{n-1}\\
&\ge e^x-\left(1+\frac xn\right)^n\\[3pt]
&\ge0
\end{align}
$$
So on $[0,1]$,
$$
0\le e^x-\left(1+\frac xn\right)^n\le e-\left(1+\frac1n\right)^n
$$
A: The logarithm $\ln:[1,e]\to[0,1]$ is uniformly continuous on this interval, so it is a 
uniform homeomorphism, and it suffices to show that $\ln f_n(x)=n\ln (1+x/n)$ is uniformly convergent on $[0,1]$.  But this follows immediately from Taylor's theorem:
$$
n\ln(1+x/n)-x=O(x^2/n).
$$
