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.

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!