Is there a way to show that $e^x=\lim_{n\to \infty }\left(1+\frac{x}{n}\right)^n$?

Question

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 ?

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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.

Answer 1

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

Answer 2

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

Answer 3

$$\frac xn(\frac n{n+x})\le\int_1^{1+\frac xn}\frac1t dt\le\frac xn(1)\implies \frac x{n+x}\le\ln (1+\frac xn)\le\frac xn\implies e^{\frac x{n+x}}\le1+\frac xn\le e^{\frac xn}\implies e^{\frac{xn}{n+x}}\le(1+\frac xn)^n\le e^x\implies e^x\le\lim_{n\to\infty}(1+\frac xn)^n\le e^x$$, by the squeeze or sandwich theorem...

Answer 4

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 :

• logarithm as an integral
• exponential/logarithm as a limit : relevant to current question
• general real power as continuous extension of rational power

Answer 5

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.