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I have to show that $$\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n=e^x$$ for all $x\in\mathbb R$ using the fact that $$e=\lim_{n\to\infty }\left(1+\frac{1}{n}\right)^n.$$

I already showed that for all $q\in\mathbb Q$, $$\lim_{n\to\infty }\left(1+\frac{q}{n}\right)^n=e^q,$$ but I have problem to show the relation for $x\in\mathbb R$. My idea is to take a sequence of rational $(x_n)$ that converge to $x\in\mathbb R\backslash \mathbb Q$ (which exist), and thus

$$\lim_{n\to\infty }\left(1+\frac{x_m}{n}\right)^n=e^{x_m}\implies \lim_{m\to\infty }\lim_{n\to\infty }\left(1+\frac{x_m}{n}\right)^n=\lim_{m\to\infty }e^{x_m},$$ but now, how can I justify that $$\lim_{m\to\infty }\lim_{n\to\infty }\left(1+\frac{x_m}{n}\right)^n=\lim_{n\to\infty }\lim_{m\to\infty }\left(1+\frac{x_m}{n}\right)^n$$ and $$\lim_{m\to\infty }e^{x_m}=e^{\lim_{m\to\infty }x_m}.$$

(I can't use continuity since I'm not supposed to know that $f:x\mapsto e^x$ is continuous).

jameselmore
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Rick
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3 Answers3

5

NOTE

We will present an outline of the proof for the case where $x\ge x_m$ for all $m$ and leave the both the details of the proof and development of the case for $x\le x_m$ to the reader.

First, we form a sequence $x_m$ for which $x_m\le x$ for all $m$, and $x_m\to x$.

Then, for any given $\epsilon>0$, there exists a number $M$ such that for all $m>M$, $0\le x-x_m <\epsilon$.

Next, we write

$$\begin{align} 0\le \left|\left(1+\frac{x}{n}\right)^n-\left(1+\frac{x_m}{n}\right)^n\right|&=\left(1+\frac{x_m}{n}\right)^n\left(\frac{\left(1+\frac{x}{n}\right)^n}{\left(1+\frac{x_m}{n}\right)^n}-1\right)\\\\ &=\left(1+\frac{x_m}{n}\right)^n\left(\frac{1}{\left(1-\frac{x-x_m}{n+x}\right)^n}-1\right) \tag 1\\\\ &\le\left(1+\frac{x_m}{n}\right)^n\left(\frac{1}{\left(1-\frac{n(x-x_m)}{n+x}\right)}-1\right) \tag 2\\\\ &=\left(1+\frac{x_m}{n}\right)^n\left(\frac{n}{n(1-(x-x_m))+x}\right)(x-x_m) \tag 3 \end{align}$$

where in going from $(1)$ to $(2)$, we used Bernoulli's Inequality.

Now, with $m>M$ held fixed, but $M$ chosen so large that $x-x_m<1$, we take $n\to \infty$ in $(3)$. Then, we have for all $\epsilon>0$ there is an $M$ so that

$$\left|\left(1+\frac{x}{n}\right)^n-\left(1+\frac{x_m}{n}\right)^n\right|\le e^{x_m}\frac{(x-x_m)}{1-(x-x_m)}<\left(\frac{e^{x_m}}{1-(x-x_m)}\right)\,\epsilon$$

and we are done!

Mark Viola
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  • thank you very much for your answer. I will let you know if I need some precision. Thanks a lot :) – Rick Oct 27 '15 at 20:28
  • Youre welcome. My pleasure. – Mark Viola Oct 27 '15 at 20:35
  • I just don't understand the last thing : We have $$\lim_{n\to\infty }|(1+x/n)^n-(1+x_m/n)^n|\leq (\frac{e^{x_m}}{1-(x-x_m)})\varepsilon,$$ Not $$(1+x/n)^n-(1+x_m/n)^n|\leq (\frac{e^{x_m}}{1-(x-x_m)})\varepsilon,$$ no ? And why having proving that we are done ? I don't see why $(1+x/n)^n\to e^x$ (I even don't understand how you define $e^x$) – user349449 Jul 06 '18 at 16:55
  • @MathBeginner Note we have $$\left(1+\frac{x_m}n\right)^n\le e^{x_m}$$AND $$\frac{n}{n(1-(x-x_m))+x}\le \frac{1}{1-(x-x_m)}$$Next, read the OP. The OP has already shown that $e^{p/q}=\lim_{n\to\infty}\left(1+\frac{p/q}{n}\right)^{n}$. With $x_m\in \mathbb{Q}$,with $x_m\to x\ge x_m$, we have shown that $e^x=\lim_{m\to\infty} e^{x_m}$ – Mark Viola Jul 06 '18 at 17:18
  • @MarkViola: So why you take $n\to \infty $ ? "Now, with m>M held fixed, but M chosen so large that x−xm<1, we take n→∞ in (3)". Moreover, I still don't understand why your very last inequality prove the claim. It's not clear, sorry. – user349449 Jul 06 '18 at 18:07
  • Well, we took the limit to arrive at the second to last inequality. The first bound I wrote actually needs to be proven. But we don't need it, we need simply to let $n\to \infty$. – Mark Viola Jul 06 '18 at 18:16
3

Step 1. If $x<y$, then $\displaystyle\left(1+\frac{x}{n}\right)^n<\left(1+\frac{y}{n}\right)^n$ - Straightforward.

Step 2. Assuming that $\,\lim_{n\to\infty}\left(1+\frac{q}{n}\right)^n=\mathrm{e}^q$, for all $q\in\mathbb Q$, then using Step 1, for every $p<x<q$, $p,q\in\mathbb Q$ and $x\in\mathbb R$, we have that $$ \mathrm{e}^p\le\liminf_{n\to\infty}\left(1+\frac{x}{n}\right)^n =:a(x) \le b(x):=\limsup_{n\to\infty}\left(1+\frac{x}{n}\right)^n \le\mathrm{e}^q. $$

Step 3. Using the fact that $f(x)=\exp(x)$ is continuous and increasing, let $x\in\mathbb R$, and let $\{p_n\}_{n\in\mathbb N},\{q_n\}_{n\in\mathbb N}\subset\mathbb Q$, $\{p_n\}_{n\in\mathbb N}$ increasing, $\{q_n\}_{n\in\mathbb N}$ decreasing, such that $p_n,q_n\to x$, then as $\exp(x)$ is continuous and increasing $\lim_{n\to\infty}\mathrm{e}^{p_n}=\lim_{n\to\infty}\mathrm{e}^{q_n}=\mathrm{e}^x$ and $$ \mathrm{e}^{p_n} < \mathrm{e}^x<\mathrm{e}^{q_n} \quad\text{and}\quad \mathrm{e}^{p_n}\le\liminf_{n\to\infty}\left(1+\frac{x}{n}\right)^n =:a(x) \le b(x):=\limsup_{n\to\infty}\left(1+\frac{x}{n}\right)^n \le\mathrm{e}^{q_n}. $$ Hence $a(x)=b(x)=\mathrm{e}^x$, and thus the limit $\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$ exists and it is equal to $\mathrm{e}^x$.

Yiorgos S. Smyrlis
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0

Just an other way

Let $f_n(x)=\left(1+\frac{x}{n}\right)^n$. The $f_n$ are continuous and converge uniformly to $e^x$ on all compact set. Therefore, $x\mapsto e^x$ is continuous on all compact set and thus

$$e^x\underset{continuity}{=}\lim_{m\to\infty }e^{x_m}=\lim_{m\to\infty }\lim_{n\to\infty }f_n(x_m)\underset{f_n\ converge\ uniformly}{=}\lim_{n\to\infty }\lim_{m\to\infty }f_n(x_m)\underset{continuity}{=}\lim_{n\to\infty }f_n(x)=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n.$$

Do you understand ?

Surb
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