## proof for exponentials

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!
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 Quote by island-boy 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 ?

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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.

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## proof for exponentials

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

 Quote by Hurkyl 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:
$$\lim_{n\rightarrow\infty}1^n + \lim_{n\rightarrow\infty} x + \lim_{n\rightarrow\infty} \frac{n(n-1)(x^2)}{(n^2)(2!)} + \lim_{n\rightarrow\infty} \frac{n(n-1)(n-2)(x^3)}{(n^3)(3!)} + \lim_{n\rightarrow\infty} \frac{n(n-1)(n-2)(n-3)(x^4)}{(n^4)(4!)} + ... = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +... =e^x$$

these correct?
 Recognitions: Gold Member Science Advisor Staff Emeritus You've used n-->infinity in expressions that don't even have an n in them! (You meant h-->0)
 Recognitions: Homework Help See this thread posts # 3 and 12. I take it that your definition for $$e^x$$ is the Taylor series about x=0, viz. $$e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$

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Essentially, this

 Quote by benorin 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,

 Quote by benorin 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 nk when the product is expanded, and hence $$\frac{n!}{(n-k)!} = n^k + \mbox{ some polynomial of degree k-1 in }n$$ or rather $$\frac{n!}{(n-k)!}\frac{1}{n^{k}} \rightarrow 1 \mbox{ as }n\rightarrow\infty$$

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