/sci/ - Science & Math » Thread #13095022

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Anonymous Fri 07 May 20:57:25 2021 No.13095022 View Reply Original Report

Quoted By: >>13095059 >>13097911

I just found out that sinhx^(1/x) approaches e at infinity.
Any algebraic reasons or what not as to why this is true?

Anonymous

Anonymous Fri 07 May 2021 20:58:45 No.13095030 Report

Quoted By:

Look up the definition of sinh and it might make sense

Anonymous

Anonymous Fri 07 May 2021 21:05:07 No.13095059 Report

Quoted By: >>13096501 >>13097352 >>13097911

>>13095022
$ \lim_{x\rightarrow \infty} \sqrt[x]{\sinh x} = \lim_{x\rightarrow \infty} \left( \frac{e^x - e^{-x}}{2} \right)^{1/x} = \frac{e^{x/x}}{2^0} = e $

Anonymous

Anonymous Sat 08 May 2021 03:43:18 No.13096501 Report

Quoted By:

>>13095059
MY GOD HE'S DONE IT
HE'S PROVED THE OIAF CONJECTURE
AND IT'S AN ELEMENTARY PROOF, NO LESS

Anonymous

Anonymous Sat 08 May 2021 04:43:11 No.13096750 Report

Quoted By: >>13097379 >>13097911

(e^x - e^-x)^(1/x)
how does that turn into e^(x/x)

Anonymous

Anonymous Sat 08 May 2021 08:44:15 No.13097352 Report

Quoted By: >>13097357

>>13095059
second equality unjustified

Anonymous

Anonymous Sat 08 May 2021 08:46:12 No.13097357 Report

Quoted By: >>13097398

>>13097352
literally the definition of sinh

Anonymous

Anonymous Sat 08 May 2021 08:54:33 No.13097379 Report

Quoted By: >>13097911

>>13096750
for x big and hard
(1/2 e^x)^(1/x) < (e^x - e^-x)^(1/x) < (e^x)^(1/x)
squeeze your butt, I'm going in dry

Anonymous

Anonymous Sat 08 May 2021 09:03:38 No.13097398 Report

Quoted By: >>13097794

>>13097357
I've said second

Anonymous

Anonymous Sat 08 May 2021 11:47:31 No.13097794 Report

Quoted By: >>13097814

>>13097398
maybe neighbourhoods of infinity are more your speed

Anonymous

Anonymous Sat 08 May 2021 12:00:15 No.13097814 Report

Quoted By: >>13097828 >>13097911

>>13097794
neighborhoods of infinity are cool, thanks for asking. now prove $\lim_{x\to\infty}\left(\frac{e^x - e^{-x}}{2} \right)^{\frac{1}{x}} = e$

Anonymous

Anonymous Sat 08 May 2021 12:04:04 No.13097828 Report

Quoted By: >>13097860 >>13099760

>>13097814
Prove it yourself, highschooler.

Anonymous

Anonymous Sat 08 May 2021 12:15:03 No.13097860 Report

Quoted By:

>>13097828
there's no shame in admitting that you can't

Anonymous

Anonymous Sat 08 May 2021 12:30:48 No.13097911 Report

Quoted By: >>13097917 >>13097919

>>13095022
>>13095059
>>13096750
>>13097379
>>13097814
$ (e^x - e^{-x})^{\frac{1}{x}} = \\ = e^{\frac{1}{x} \log(e^x - e^{-x})} \\ = e^{\frac{1}{x} \log(e^x (1 - e^{-2x})} \\ = e^{\frac{1}{x} (x + log(1 - e^{-2x})} \\ = e (1 - e^{-2x}) \to e $

Anonymous

Anonymous Sat 08 May 2021 12:35:12 No.13097917 Report

Quoted By:

>>13097911
hmmm last step I messed up,
e^[(1/x) log(1 - e^(-2x)] --> e^[0 * log(1-0)] = 1

Anonymous

Anonymous Sat 08 May 2021 12:38:03 No.13097919 Report

Quoted By:

>>13097911
sinhx = (e^x - e^-x)/2

Anonymous

Anonymous Sat 08 May 2021 20:39:35 No.13099760 Report

Quoted By:

>>13097828
Cope

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