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DLSBest ResponseYou've already chosen the best response.2
\[\LARGE \lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{x} e}{x}\]
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Taylor Series expansion?
 11 months ago

experimentXBest ResponseYou've already chosen the best response.0
Taylor series works
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Weird. The Taylor Series expansion is in the terms of \(e\) and \(x\). The first term is \(e\)
 11 months ago

experimentXBest ResponseYou've already chosen the best response.0
gives e/2 if you calculate correctly
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Oh wait! lol, yeah. I forgot to consider that e/2 which had no \(x\)
 11 months ago

DLSBest ResponseYou've already chosen the best response.2
How about a different way rather than taylor series?
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Definition of derivative? :
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
I was just thinking how this might be the definition of a derivative in disguise.
 11 months ago

experimentXBest ResponseYou've already chosen the best response.0
Binomial expansion might work ... but that's too long
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
DLS Use whatever you want. 15 minutes ago DLS How about a different way rather than taylor series? 9 minutes ago
 11 months ago

DLSBest ResponseYou've already chosen the best response.2
You are right,still investigating for more methods,I wrote so to specify that you are not bound to any particular method,just asking if you know anything alternate :)
 11 months ago

experimentXBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=Limit%5B%281+%2B+x%29%5E%281%2Fx%29+%281%2F%28x+%281+%2B+x%29%29++Log%5B1+%2B+x%5D%2Fx%5E2%29%2C+x+%3E+0%5D after using L'hopital you get this expression. The first term tends to e .. if you again apply L'hopital on the second term, you will get 1/2
 11 months ago
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