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/sci/ - Science & Math

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5791574 No.5791574 [Reply] [Original]

How do I prove that <span class="math">{(1 + \frac{x}{n})}^n[/spoiler] converges to <span class="math">e^x[/spoiler]?
I cannot into series.

>> No.5791582

by using A=e^ln(A), then its essentially just proving ln((1+x/n)^n) converges to x

>> No.5791590

>>5791582
And how would I prove that?
<span class="math">lim(ln((1 + \frac{x}{n})^n)) = ln((1 + \frac{x}{\infinity})^\infinity) = ln((1 + 0)^\infinity) = ln(1^\infinity) = ln(1)[/spoiler]

What.

>> No.5791592

>>5791582
And how would I prove that?
<span class="math">lim(ln((1 + \frac{x}{n})^n)) = ln((1 + \frac{x}{infinity})^infinity) = ln((1 + 0)^infinity) = ln(1^infinity) = ln(1)[/spoiler]

What.

>> No.5791594

>>5791592
>>5791590

Fuck you, LaTeX.
<span class="math">lim(ln((1 + \frac{x}{n})^n)) = ln((1 + \frac{x}{infinity})^{infinity}) = ln((1 + 0)^{infinity}) = ln(1^{infinity}) = ln(1)[/spoiler]

What.

>> No.5791595

Since you know (1 + 1/n)^n converges to e, you can make a substitution m = n/x

Now, since m linear with n and x not part of the limit, your limit is as m goes to infinity of (1 + 1/m)^(mx)

The limit is only of m, so the x exponent can be taken out of the limit:
[lim m ->inf (1 + 1/m)^m]^x
which is precisely e^x, since the inner limit goes to e for any sequence going to infinity (m here).

>> No.5791602

>>5791594
1^infinity is indeterminate, you actually have to do a proper limit. I mean by your logic you could say e^x=1.

I'll give you two hints. First, ln((1+x/n)^n)=n*ln(1+x/n).
Second, is l'hopitols with n=1/(1/n)

>> No.5791601

>>5791595
Nice. Thank you.

>> No.5791616

>>5791574
http://www.wolframalpha.com/input/?i=lim+n-%3Einfinity+%281%2B%28x%2Fn%29%29%5En

Click "step-by-step solution".

>> No.5791642

That's the definition of e though... That's like asking, "how can I prove pi is the ratio of a circles circumference to its diameter?"

>> No.5791679

>>5791642
> not using 1 + x + x^2/2 + x^3/6 + .. + x^/n! + ...