EddieDoesMath
Region: US
Friday 08 May 2026 03:33:13 GMT
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فيصل 🇦🇪🇱🇧🇵🇸 :
using taylor series while finding the derivative of e^x is circular reasoning as you need the derivative of e^x to find the taylor series expansion of e^x
2026-05-24 18:02:41
5
Safi :
your proof is wrong since using taylor series presumes the derivative of e^x is e^x
2026-05-09 14:11:29
236
haard.climbs :
This is circular, the derivative is used to get the Taylor expansion of e^h. Therefore the only way to prove this is to use other methods to evaluate the limit. And no, you can’t use L’Hospital’s rule either.
2026-05-08 06:22:17
269
jonny21110 :
This is thé worst proof I’ve ever seen. You can’t do this.
2026-05-09 20:05:25
27
marxslop :
d/dx[e^x] = xe^x-1 😅
2026-05-09 05:42:07
77
---- :
this is completely circular since the Taylor series of f(x)=e^x is Σ[f'(x-a)/n!]*(x-a)^n. you literally can't derive the Taylor series of e^x without already knowing the derivative of e^c
2026-05-12 22:53:07
0
Jonathan :
but that proof has nothing to do with 2.718....., you are just using e as a variabel
2026-05-26 16:43:14
3
vespero :
oh boy, circular demonstration.
2026-05-29 20:00:53
6
Anouar :
Is it because: f’(x) = e^x * ln(e) = e^x
2026-05-09 11:52:45
13
Bela :
I thought it was because, a number that approaches 0 divided by a number that approaches 0 is just 1, because any number divided by itself is 1.
2026-05-26 05:48:15
0
Mejiarma :
As h reaches 0, limit = [ ( e^x - e^x ) / h ] = 0/h = 0
2026-05-10 15:03:16
0
Le Wronskien :
😅and of course Taylor series come from derivatives so your proof it's quite circular
2026-05-09 16:50:50
7
poppi256 :
If you use Taylor, then you could have aplied it at first hand. You dont need even the limit. just derive the Taylor series
2026-05-09 17:02:21
2
King_Chlamydia :
Crazy nice graph there with zero effort😭
2026-05-11 06:20:49
1
Owaes Atme :
you can use tylor to prove that cause you dont now tylor for e if you dont now the derivative
2026-05-18 09:38:04
1
romariohx :
That’s not a proof . The derivative of e^x is e^x cause we defined it that way , trying to find the derivative of a^x . 😉😉
2026-05-10 01:27:36
2
😴 :
The taylor series require the derivative of e^x. Also it is inappropriate to show that the series 1 + h + h^2/2! + … approach 1 wothout a proper proof, since it is the sum of infinite number of terms
2026-05-10 07:11:08
2
Janaya ✝️ :
circular reasoning mate
2026-05-09 08:13:47
6
emmanueloppong886 :
i love how you try to explain from basic concept.
2026-05-11 14:19:36
1
Gold3nn :
what is h for?
2026-05-10 19:27:02
1
poppi256 :
just proof it by d/dx of the Euler or Bernoulli formula (idk whats the name) lim((1+x/n)^n) :
2026-05-18 16:39:43
1
Joe.Hunting :
Wrong, It’s circular reasoning.
2026-05-10 08:32:24
1
roadsahead23 :
I have no idea how Taylor series works imma just take ur word it makes sense 😂
2026-05-19 05:57:37
1
Cours_En_Ligne_CORAN_ARABE :
thank so much 🥰
2026-05-09 15:57:07
1
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