Prove that this is e^x

**Bruno J.** · August 9th 2009, 04:16 PM

Prove the famous identity

$\lim_{n\rightarrow \infty} \left(1+\frac{x}{n}\right)^n = e^x$

**Random Variable** · August 9th 2009, 04:36 PM

let $y = \lim_{n \to \infty} \Big(1+\frac{x}{n}\Big)^{n}$

$\ln y = \lim_{n \to \infty} \ln \Big[\Big(1+\frac{x}{n}\Big)^{n}\Big]$

$= \lim_{n \to \infty} n \ln \Big[\Big(1+\frac{x}{n}\Big)\Big]$

$= \lim_{n \to \infty} \frac{\ln \Big[\Big(1+\frac{x}{n}\Big)\Big]}{\frac{1}{n}}$

$= \lim_{n \to \infty} \frac{\frac{1}{1+\frac{x}{n}}\frac{-x}{n^{2}}}{\frac{-1}{n^2}}$

$= \lim_{n \to \infty} \frac{x}{1+\frac{x}{n}} = x$

so if $\lim_{n \to \infty} \ln y = x$ then $\lim_{n \to \infty} y = e^{x}$

**Bruno J.** · August 9th 2009, 05:30 PM

Good!
Let's see in how many different ways we can prove it!

Here is my proof, without l'Hôpitals; let $P_n = \left(1-\frac{x}{n}\right)^n$ . Then

$\log P_n = n\log\left(1-\frac{x}{n}\right)$ .
As soon as $n>|x|$ , we can use the series expansion $\log(1-y)=-\sum_{k=1}^\infty\frac{y^k}{k}$ with $y=\frac{x}{n}$ to get

$\log P_n = -n\sum_{k=1}^\infty\frac{x^k}{kn^k} = -\sum_{k=1}^\infty\frac{x^k}{kn^{k-1}}$

$= -x\sum_{k=1}^\infty\frac{x^{k-1}}{kn^{k-1}}=-x\left(1+\sum_{k=2}^\infty\frac{x^{k-1}}{kn^{k-1}}\right)$

and $\left|\sum_{k=2}^\infty\frac{x^{k-1}}{kn^{k-1}}\right| \leq \sum_{k=2}^\infty\frac{|x|^{k-1}}{kn^{k-1}} \leq \sum_{k=2}^\infty\frac{|x|^{k-1}}{n^{k-1}} = \frac{|x|/n}{1-|x|/n} = \frac{|x|}{n-|x|} \rightarrow 0$ as $n \rightarrow \infty$ .

So as $n \rightarrow \infty$ , we have $\log P_n = -x\left(1+\sum_{k=2}^\infty\frac{x^{k-1}}{kn^{k-1}}\right) \rightarrow -x$ , so $P_n \rightarrow e^{-x}$ .

**Moo** · August 9th 2009, 11:38 PM

Hello,

$\ln P_n=n \ln\left(1+\frac xn\right)$

As n goes to infinity, $\frac xn$ goes to 0.

Thus we have the asymptotic equivalent : $\ln\left(1+\frac xn\right)=\frac xn +O\left(\frac 1n\right)$

Then $\ln P_n=x+O(1)$

---> $\lim_{n\to\infty} \ln P_n=x$

And it follows that $\lim_{n\to\infty}P_n=e^x$

**Bruno J.** · August 10th 2009, 07:13 PM

Thank you Moo, that is much simpler.

Here is another proof I thought of today. Taking the logarithmic derivative of $Q_n(z)=\left(1+\frac{z}{n}\right)^n$ , we get $n\left(\frac{1/n}{\left(1+\frac{z}{n}\right)}\right)=\frac{1}{\le ft(1+\frac{z}{n}\right)}$ , and this goes to $1$ as $n\rightarrow \infty$ . Hence the function $f(z)=\lim_{n\rightarrow \infty}Q_n(z)$ satisfies $\frac{f'(z)}{f(z)}=1$ for all complex $z$ , and it follows that $f(z)=e^z$

Edit : oops, this is exactly what Random Variable did

**Failure** · August 10th 2009, 09:22 PM

Originally Posted by Bruno J.

Prove the famous identity

$\lim_{n\rightarrow \infty} \left(1+\frac{x}{n}\right)^n = e^x$

"To prove" means to argue for the validity of a statement based on prior knowledge. But what prior knowledge are we allowed to presuppose? In introductory courses, for example, this limit is sometimes used to define $\mathrm{e}^x$ . Therefore, no prior knowledge of this function or its inverse may be used to show the existence of the limit in such a case.

**Laurent** · September 21st 2009, 06:34 AM

Originally Posted by Bruno J.

Prove the famous identity

$\lim_{n\rightarrow \infty} \left(1+\frac{x}{n}\right)^n = e^x$

Instead of involving logarithm and its development in the neighbourhood of 1, we can use (in a more "algebraic" way) the power series definition of the exponential: $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$ .

We develop $\left(1+\frac xn\right)^n=\sum_{k=0}^n{n\choose k}x^k$ $=\sum_{k=0}^n \frac{x^k}{k!}\frac{n(n-1)\cdots (n-k+1)}{n^k}$ .

For a fixed $k$ , the ratio $\frac{n(n-1)\cdots (n-k+1)}{n^k}$ tends to 1 (the numerator is a monic polynomial of degree $k$ in the variable $n$ ).

"Formally", we thus have $\left(1+\frac xn\right)^n\to_n \sum_{k=0}^\infty \frac{x^k}{k!}=e^x$ .

The good point of this proof is that we can "see" why the limit is the exponential; this may have been Euler's way to prove the formula. However, it is not trivial to prove the limit rigorously (there is a summation in $k$ , not a fixed index...). This can probably be done elementarily, but here is a short not-so-elementary way. We have $\sum_{k=0}^n {n\choose k}x^k=\sum_{k=0}^\infty a_{k,n}$ where $a_{k,n}=0$ for $k>n$ , and $a_{k,n}={n\choose k}x^k$ otherwise. For fixed $k$ , $a_{k,n}\to_n \frac{x^k}{k!}$ . The previous ratio is bounded by 1, hence $|a_{k,n}|\leq \frac{|x|^k}{k!}$ for all $k,n$ . We have $\sum_{k=0}^\infty \frac{|x|^k}{k!}<\infty$ , therefore we can apply the bounded convergence theorem to take the limit in $n$ in the summation.

Originally Posted by Moo

Thus we have the asymptotic equivalent : $\ln\left(1+\frac xn\right)=\frac xn +{\color{red}O}\left(\frac 1n\right)$

Then $\ln P_n=x+{\color{red}O}(1)$

---> $\lim_{n\to\infty} \ln P_n=x$

En passant: you probably meant little o's. Indeed, $x+O(1)$ is a bounded sequence, and not a sequence converging toward $x$ (contrary to $x+o(1)$ ). The first two lines were correct, but you couldn't deduce the limit with big O's.

**Prove It** · September 21st 2009, 09:33 AM

Originally Posted by Bruno J.

Prove the famous identity

$\lim_{n\rightarrow \infty} \left(1+\frac{x}{n}\right)^n = e^x$

If you use a function $f(x) = a^x$ and try to find its derivative, by defining $e$ as the value of $a$ that makes $f'(x) = f(x)$ we have...

$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$

$= \lim_{h \to 0}\frac{a^{x + h} - a^x}{h}$

$= \lim_{h \to 0}\frac{a^x(a^h - 1)}{h}$

$= a^x \lim_{h \to 0}\frac{a^h - 1}{h}$ .

For the function to be its own derivative, we would require that

$\lim_{h \to 0}\frac{a^h - 1}{h} = 1$

and for simplicity, we will now use the symbol $e$ to represent this particular value of $a$ .

So we have

$\lim_{h \to 0}\frac{e^h - 1}{h} = 1$

$\lim_{h \to 0}(e^h - 1) = \lim_{h \to 0} h$

$\left[\lim_{h \to 0}e^h\right] - 1 = \lim_{h \to 0} h$

$\lim_{h \to 0}e^h = 1 + \lim_{h \to 0}h$

$\lim_{h \to 0}e^h = \lim_{h \to 0}\left(1 + h\right)$

$e = \lim_{h \to 0}(1 + h)^\frac{1}{h}$ .

Now by letting $n = \frac{1}{h}$ , we notice that as $h \to 0, n \to \infty$ .

So $e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$

Can you go from here?

**bkarpuz** · September 21st 2009, 09:40 AM

Originally Posted by Prove It

So we have
$\lim_{h \to 0}\frac{e^h - 1}{h} = 1$
$\lim_{h \to 0}(e^h - 1) = \lim_{h \to 0} h$ (*)
$\left[\lim_{h \to 0}e^h\right] - 1 = \lim_{h \to 0} h$
...
$\lim_{h \to 0}e^h = 1 + \lim_{h \to 0}h$
$\lim_{h \to 0}e^h = \lim_{h \to 0}\left(1 + h\right)$
$e = \lim_{h \to 0}(1 + h)^\frac{1}{h}$ . (**)

But you have $0/0$ at the line above of (*) I mark above, I think you cant do that.
Also you cant pass to (**) i think...

**Prove It** · September 21st 2009, 09:44 AM

Originally Posted by bkarpuz

But you have $0/0$ at the point (*) I mark above, I think you cant do that.

There is nothing wrong with what I have written.

Indeterminate limits are common and CAN be used - I have seen it be done in numerous texts.

Also, the rules of exponentiation hold even when taking limits.

**bkarpuz** · September 21st 2009, 09:52 AM

Originally Posted by Prove It

There is nothing wrong with what I have written.

Indeterminate limits are common and CAN be used - I have seen it be done in numerous texts.

Also, the rules of exponentiation hold even when taking limits.

But if you separate the limits then you may change the speeds of the variables, i.e., you may use $\lim_{h\to0}h^{2}$ instead of $\lim_{h\to0}h$ , which will lead you to a different result...

Also in (**), you can not take the power $\cdot^{h}$ outside the limit either since it is not a constant, i.e., depends on the limit variable...

A simple note.
$\lim_{n\to0}\frac{f(n)}{g(n)}=\frac{\lim\limits_{n \to0}f(n)}{\lim\limits_{n\to0}g(n)}$ provided that $\lim_{n\to0}g(n)\neq0$ .

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