How Does the Limit of (1 + x/n)^n as n Approaches Infinity Equal e^x?

[island-boy]

hello, I need the proof to show that:
$$e^x = \lim_{n\rightarrow\infty} (1+x/n)^n$$

Here's what I was able to come up with so far:
since the derivative of $$e^x$$ is also $$e^x$$,

then let f(x) = $$e^x$$
thus:
D(f(x)) = $$\lim_{n\rightarrow\infty} \frac{f(x+h) - f(x)}{h} = \lim_{n\rightarrow\infty} \frac{e^{x+h} - e^x}{h} = e^x\lim_{n\rightarrow\infty} \frac{e^h - 1}{h}$$

so for the derivative of $$e^x$$ to equal itself,
$$\lim_{n\rightarrow\infty} \frac{e^h - 1}{h} = 1$$

so for small values of h, we can write:
$$e^h - 1 = h$$
and so
$$e = (1+h)^{1/h}$$

Replacing h by 1/n, we get:
$$e = (1 + 1/n)^n$$
As n gets larger and approaches infinity, we get:
$$e = \lim_{n\rightarrow\infty} (1+1/n)^n$$

so, how do I get:
$$e^x = \lim_{n\rightarrow\infty} (1+x/n)^n$$

Also, is it true that:
$$(1+x/n)^n \leq e^x$$ and
$$(1-x/n)^n \leq e^{-x}$$ for every natural n and every x element of X?
How can I prove this?

thanks!

 

[Fermat]

hello, I need the proof to show that:
$$e^x = \lim_{n\rightarrow\infty} (1+x/n)^n$$
...
thanks!

Have you tried the Binomial theorem ?

 

[Hurkyl]

I would've taken the expression for e and raised both sides to the x power.

so for small values of h, we can write:

You really ought to do this more rigorously...

P.S. many of the limits in your limits are wrong.

 

[arildno]

As a hint, set 1/u=x/n, and re-express in terms of u and x.

 

[island-boy]

I would've taken the expression for e and raised both sides to the x power.You really ought to do this more rigorously...

P.S. many of the limits in your limits are wrong.

which one of the limits are wrong?

I was able to prove that:
$$(1 + x/n)^n \leq e^x$$ and $$(1 - x/n)^n \leq e^{-x}$$

Using the Taylor expansion for $$e^x$$, I got

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
which means
$$e^x \geq 1 + x$$
subtituting x by x/n, I get
$$e^{x/n} \geq 1 + x/n$$
thus
$$e^{(x/n)n} \geq (1 + x/n)^n$$
simplifying, I get:
$$e^x \geq (1 + x/n)^n$$

Similarly I used the Taylor Expnansion to prove that
$$e^x = \lim_{n\rightarrow\infty} (1+x/n)^n$$

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

and using binomial theorem for $$\lim_{n\rightarrow\infty} (1+x/n)^n$$
I got:
$$\lim_{n\rightarrow\infty}1^n + \lim_{n\rightarrow\infty} x + \lim_{n\rightarrow\infty} \frac{n(n-1)(x^2)}{(n^2)(2!)} + \lim_{n\rightarrow\infty} \frac{n(n-1)(n-2)(x^3)}{(n^3)(3!)} + \lim_{n\rightarrow\infty} \frac{n(n-1)(n-2)(n-3)(x^4)}{(n^4)(4!)} + ...<br /> <br /> = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +...<br /> <br /> =e^x$$

these correct?

 

[Hurkyl]

You've used n-->infinity in expressions that don't even have an n in them! (You meant h-->0)

 

[benorin]

See this thread posts # 3 and 12.

I take it that your definition for $$e^x$$ is the Taylor series about x=0, viz.

$$e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$

 

[benorin]

Essentially, this

That $$\lim_{n\rightarrow\infty} (1+\frac{x}{n})^n = e^x$$ may be proven as follows:

by the binomial theorem,

$$(1+\frac{x}{n})^n = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!}\left( \frac{x}{n}\right) ^{k} = \sum_{k=0}^{n} \frac{n!}{(n-k)!}\frac{1}{n^{k}} \frac{x^k}{k!}\rightarrow \sum_{k=0}^{\infty} \frac{x^k}{k!}=:e^{x}\mbox{ as }n\rightarrow\infty$$

since $$\frac{n!}{(n-k)!}\frac{1}{n^{k}} \rightarrow 1 \mbox{ as }n\rightarrow\infty$$

although there are some uniform convergence issues to be handled when taking the limit of the above sum...

and this,

This is not rigorous, but it works...

$$\frac{n!}{(n-k)!} = \frac{n(n-1)(n-2)\cdots (n-(k-1))(n-k)!}{(n-k)!} =n(n-1)(n-2)\cdots (n-(k-1))$$

counting the number of terms in the above product: it goes n-0 through n-(k-1) so there are (k-1)-0+1 = k terms so we know that the leading term will be n^k when the product is expanded, and hence

$$\frac{n!}{(n-k)!} = n^k + \mbox{ some polynomial of degree k-1 in }n$$

or rather

$$\frac{n!}{(n-k)!}\frac{1}{n^{k}} \rightarrow 1 \mbox{ as }n\rightarrow\infty$$