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I know that $$e:=\lim_{n\to \infty }\left(1+\frac{1}{n}\right)^n,$$ by definition. Knowing that, I proved successively that $$e^{k}=\lim_{n\to \infty }\left(1+\frac{k}{n}\right)^n,$$ when $k\in \mathbb N$, $k\in \mathbb Z$ and $k\in\mathbb Q$. Now, I was wondering : how can I extend this result over $\mathbb R$ ? I tried to prove that $f_n(x):=(1+\frac{x}{n})^n$ converge uniformly on $\mathbb R$ but unfortunately it failed (I'm not sure that it's even true). Any idea ?

My idea was to define the function $x\longmapsto e^x$ as $$e^x=\begin{cases}e^x& x\in \mathbb Q\\ \lim_{n\to \infty }e^{k_n}&\text{if }k_n\to x \text{ and }(k_n)\subset \mathbb Q\end{cases}.$$ But to conclude that $$e^x=\lim_{n\to \infty }\left(1+\frac{x}{n}\right)^n,$$ I need to prove that $f_n(x)=\left(1+\frac{x}{n}\right)^n$ converge uniformly on a neighborhood of $x$, but I can't do it. I set $$g_n(x)=f_n(x)-e^x,$$ but I can't find the maximum on a compact that contain $x$, and thus can't conclude.

user349449
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5 Answers5

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It is not difficult to prove the result for real irrational $x$ if you have already proved the case for rational $x$. The only idea you need to establish first as a part of your definition of $e^x$ is that $f(x) =e^x$ is continuous everywhere. I leave this as an exercise for you (hint: show that $\lim_{x\to 0}e^x=1$ using your definition).

Now let $x$ be any irrational number. Given any $\epsilon>0$ there is $\delta>0$ such that $$e^x-\epsilon<e^t<e^x+\epsilon$$ whenever $|t-x|<\delta$. Consider two rationals $r, s$ with $x-\delta<r<x<s<x+\delta$ and then we have $$e^x-\epsilon <e^r<e^s<e^x+\epsilon$$ Now we have $$\left(1+\frac{r}{n}\right)^n<\left(1+\frac{x}{n}\right)^n<\left(1+\frac{s}{n}\right)^n$$ and taking limits as $n\to\infty$ we get $$e^x-\epsilon<e^r\leq \lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n\leq e^s<e^x+\epsilon $$ (the above assumes that the limit in question exists for irrational $x$ also and you can prove it using the fact that a bounded monotone sequence is convergent, or better apply liminf/limsup to the above inequalities). Since $\epsilon$ is arbitrary it follows that $$e^x=\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n$$

Based on feedback from Mark Viola via comments I am giving a link to my blog posts which discuss various routes to the theory of exponential and logarithmic functions :

Paramanand Singh
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We can use that exists $p_n, q_n \in \mathbb{Q}$ such that $p_n,q_n \to x$ and $p_n\le x\le q_n$, therefore

$$\left(1+\frac{p_n}{n}\right)^n\le \left(1+\frac{x}{n}\right)^n\le \left(1+\frac{q_n}{n}\right)^n$$

and

$$\left(1+\frac{p_n}{n}\right)^n=\left[\left(1+\frac{p_n}{n}\right)^\frac{n}{p_n}\right]^{p_n}\to e^x$$

$$\left(1+\frac{q_n}{n}\right)^n=\left[\left(1+\frac{q_n}{n}\right)^\frac{n}{q_n}\right]^{q_n}\to e^x$$

indeed for $\frac{n}{p_n}\in (m,m+1)$ with $m\in \mathbb{N}$ we have

$$\left(1+\frac1{m+1}\right)^m\le \left(1+\frac{p_n}{n}\right)^\frac{n}{p_n}\le \left(1+\frac1m\right)^{m+1}$$

and therefore $\left(1+\frac{p_n}{n}\right)^\frac{n}{p_n}\to e$.

user
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  • I think this is by far the simplest approach, and it does not rely on other properties of the exponential (e.g. its derivative or the properties of the logarithm as its inverse). The key observation is basically that $x$ can be moved from the inside of the parentheses to the exponent by substituting the variable used in the limit. – sasquires Jul 06 '18 at 16:35
  • Sorry but why $\left(\left(1+\frac{p_n}{n}\right)^{\frac{p_n}{n}}\right)^{p_n}\to e^x$ ? – user349449 Jul 06 '18 at 16:37
  • @MathBeginner since $p_n/n \to 0$ we have $$\left(1+\frac{p_n}{n}\right)^{\color{red}{\frac{n}{p_n}}}\to e$$ and $p_n \to x$. – user Jul 06 '18 at 16:39
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    Since the exponent depend on $n$, we can't do that, no ? – user349449 Jul 06 '18 at 16:43
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    @MathBeginner Yes we can since $p_n/n \to 0$. – user Jul 06 '18 at 17:12
  • @MarkViola I've assumed that the limit has been proved for $p_n/n \to 0$ with $p_n/n \in \mathbb{Q}$. – user Jul 06 '18 at 17:13
  • The problem mentioned by @MarkViola is genuine and I don't see a way around it. Instead I have used the continuity of the function $e^x$ in my answer. – Paramanand Singh Jul 06 '18 at 17:22
  • @gimusi It took me a while to realize the problem, but the problem is that the OP has only proven the limit for fixed rational $k$, not the case where $k$ depends on $n$. – sasquires Jul 06 '18 at 17:22
  • @ParamanandSingh Yes of course, I just realized that! I will revise to improve that part and of course I'll take alook to your derivation. Thanks – user Jul 06 '18 at 17:24
  • @sasquires Yes you are completely right! – user Jul 06 '18 at 17:25
  • @MarkViola I've updated with the derivation of the limit for $p_n/n \in \mathbb{Q}$. – user Jul 06 '18 at 18:30
  • @ParamanandSingh Now it should be complete! – user Jul 06 '18 at 18:31
  • @sasquires Now it should be complete. – user Jul 06 '18 at 18:31
  • @gimusi I agree. I like this approach. To me, the fundamental reason that the exponential behaves this way is because you can rescale the argument of the limit. The naive argument (whose flaws we have explored here) is $u=n/x$, so $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n=\lim_{u\to\infty}\left[\left(1+\frac{1}{u}\right)^{ux}\right]=\left[\lim_{u\to\infty}\left(1+\frac{1}{u}\right)^u\right]^x=e^x$$No other special properties of the exponential need to be invoked. Among the arguments presented here, this one most closely follows the naive argument, while being formally correct. – sasquires Jul 06 '18 at 18:41
  • @sasquires Yes by that approach once we have shown that $\lim_{u\to\infty}\left(1+\frac{1}{u}\right)^u=e$ also for $u \in Q$ or $R$, which is trivial, we can conclude directly. – user Jul 06 '18 at 18:46
  • Good job fixing a few things. I think you still need to invoke continuity because you can now do this for a fixed $p_n$ (but not for one that depends on $n$). The details depend on your definition of $e^x$, so it is difficult to tell exactly what should be added. – Jyrki Lahtonen Jul 06 '18 at 19:03
  • @JyrkiLahtonen Thanks for your kind support and suggestion! I'll try to improve further the approach. Thanks again – user Jul 06 '18 at 19:08
  • @JyrkiLahtonen He has omitted details. I will attempt to fill them in. (1) Write $m_n$ instead of $m$, where $m_n$ is that it is the integer such that $n/p_n \in [m, m+1)$. (2) Show that as $n \to \infty$, $m_n \to \infty$, which is basically trivial using the definition above because $p_n \to x$. (3) Take the limit $n \to \infty$ in the final inequality, where the left and right sides of the final inequality are a subsequence of a converging sequence, and use the sandwich inequality. This shows $$\left( 1 + \frac{p_n}{n} \right)^{n/p_n} \to e$$ and the rest follows. – sasquires Jul 06 '18 at 19:28
  • @sasquires Yes of course $m$ depends on $n$ and $m_n \to \infty$. – user Jul 06 '18 at 19:34
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To prove $$e^x=\lim_{n\to \infty }\left(1+\frac{x}{n}\right)^n$$

Let $$ y=\left(1+\frac{x}{n}\right)^n$$

$$ \ln y=n \ln(1+x/n)$$

$$= \frac {\ln(1+x/n)}{(1/n)}$$

$$\lim_{n\to \infty }\ln y=\lim_{n\to \infty }\frac {\ln(1+x/n)}{(1/n)}=x$$

Thus $$\lim_{n\to \infty } y= e^x$$

Mohammad Riazi-Kermani
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    This doesn't address the question, which is "How does one show that $$\left(\lim_{n\to\infty}\left(1+\frac1n\right)^n\right)^x =\lim_{n\to \infty}\left(1+\frac xn\right)^n?"$$ – Mark Viola Jul 06 '18 at 16:29
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$$\frac xn(\frac n{n+x})\le\int_1^{1+\frac xn}\frac1t dt\le\frac xn(1)\implies \frac x{n+x}\le\ln (1+\frac xn)\le\frac xn\implies e^{\frac x{n+x}}\le1+\frac xn\le e^{\frac xn}\implies e^{\frac{xn}{n+x}}\le(1+\frac xn)^n\le e^x\implies e^x\le\lim_{n\to\infty}(1+\frac xn)^n\le e^x$$, by the squeeze or sandwich theorem...

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Left side:

The exponential function may be written as a Taylor series:

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$

Right side:

$(1+\frac{x}{n})^n$ is a binomial expansion like:

$(1+y)^n=\binom{n}{0}y^0+\binom{n}{1}y^1+\binom{n}{2}y^2+...+\binom{n}{n-1}y^{n-1}+\binom{n}{n}y^n$

Where $\binom{n}{k}$ is the Binomial coefficient given by the formula : $\binom{n}{k}=\frac{n!}{k!(n-k)!}$

Some basic properties of $\binom{n}{k}$:

a)$\binom{n}{0}=1$ because $\frac{n!}{0!(n-0)!}=\frac{n!}{1*n!}$

b)$\binom{n}{1}=n$ because $\frac{n!}{1!(n-1)!}=\frac{(n-1)!*n}{(n-1)!}$

c)$\binom{n}{n-1}=n$ because $\frac{n!}{(n-1)!(n-(n-1))!}=\frac{(n-1)!*n}{(n-1)!*1!}$

d)$\binom{n}{n}=1$ because $\frac{n!}{n!(n-n)!}=\frac{1}{1!}$

e) The formula does exhibit a symmetry that is less evident from the multiplicative formula: $\binom{n}{k}=\binom{n}{n-k}$

Returning:

$(1+\frac{x}{n})^n=1+n*\frac{x}{n}+\frac{n!}{2!(n-2)!}\frac{x^2}{n^2}+\frac{n!}{3!(n-3)!}\frac{x^3}{n^3}+...+\frac{n!}{3!(n-3)!}\frac{x^{n-3}}{n^{n-3}}+\frac{n!}{2!(n-2)!}\frac{x^{n-2}}{n^{n-2}}+n*\frac{x^{n-1}}{n^{n-1}}+\frac{x^n}{n^n}$

$(1+\frac{x}{n})^n=1+x+\frac{(n-1)n}{n^2}\frac{x^2}{2!}+\frac{(n-2)(n-1)n}{n^3}\frac{x^3}{3!}+...+\frac{(n-2)(n-1)n}{3!}\frac{x^{n-3}}{n^{n-3}}+\frac{(n-1)n}{2!}\frac{x^{n-2}}{n^{n-2}}+\frac{x^{n-1}}{n^{n-2}}+\frac{x^n}{n^n}$

$(1+\frac{x}{n})^n=1+x+\frac{n-1}{n}\frac{x^2}{2!}+\frac{(n-2)(n-1)}{n^2}\frac{x^3}{3!}+...+\frac{(n-2)(n-1)}{n^{n-4}}\frac{x^{n-3}}{3!}+\frac{n-1}{n^{n-3}}\frac{x^{n-2}}{2!}+\frac{x^{n-1}}{n^{n-2}}+\frac{x^n}{n^n}$

Let's analyze what happens for $n\rightarrow\infty$-here we have three types of limits:

-First type:

$\displaystyle\lim_{n \to \infty}\frac{n-1}{n}=\displaystyle\lim_{n \to \infty}[1+\frac{1}{n}]=1+0=1$ $\displaystyle\lim_{n \to \infty}\frac{(n-2)(n-1)}{n^2}=\displaystyle\lim_{n \to \infty}\frac{n^2-3n+2}{n^2}=\displaystyle\lim_{n \to \infty}[1-\frac{3}{n}+\frac{2}{n^2}]=1-0+0=1$

Hense $\displaystyle\lim_{n \to \infty}\Bigg(\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^k}\Bigg)=1$

-Second type is $\displaystyle\lim_{n \to \infty} \frac{x^{n-\alpha}}{n^{n-\beta}}$-Because ${n^{n-\beta}}$ grows much fasten than $x^{n-\alpha}$ hense: $\displaystyle\lim_{n \to \infty} \frac{x^{n-\alpha}}{n^{n-\beta}}=0$

-Third type:

$\displaystyle\lim_{n \to \infty}\Bigg(\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^{n-k-1}}\frac{x^{n-k}}{k!}\Bigg)$

We have to show on the biggest power (similar to the first type) as the most relevant:

$\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^{n-k-1}}\frac{x^{n-k}}{k!}\sim\frac{n^{k-1} }{n^{n-k-1}}\frac{x^{n-k}}{k!}=n^{k-1-(n-k-1)}\frac{x^{n-k}}{k!}=n^{2k-n}*\frac{x^{n-k}}{k!}=\frac{1}{k!}*\frac{x^{n-k}}{n^{n-2k}}$

Again: ${n^{n-\beta}}$ grows much faster than $x^{n-\alpha}$

Hense: $\displaystyle\lim_{n \to \infty}\Bigg(\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^{n-k-1}}\frac{x^{n-k}}{k!}\Bigg)=0$

Our right side equals:

$\displaystyle\lim_{n \to \infty}(1+\frac{x}{n})^n=1+x+1*\frac{x^2}{2!}+1*\frac{x^3}{3!}+...+0+0+0+0$

$\displaystyle\lim_{n \to \infty}(1+\frac{x}{n})^n=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$

We got the same elements like in the Taylor series of $e^x$. Q.E.D.

Community
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Krzysztof Myśliwiec
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