DLS
  • DLS
Limits question [Challenge]
Mathematics
chestercat
  • chestercat
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DLS
  • DLS
\[\LARGE \lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{x} -e}{x}\]
DLS
  • DLS
@yrelhan4 @shubhamsrg
shubhamsrg
  • shubhamsrg
* :)

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DLS
  • DLS
first time in life :O
ParthKohli
  • ParthKohli
Taylor Series expansion?
DLS
  • DLS
Use whatever you want.
ParthKohli
  • ParthKohli
Ans 0?
DLS
  • DLS
Nope
experimentX
  • experimentX
Taylor series works
ParthKohli
  • ParthKohli
Weird. The Taylor Series expansion is in the terms of \(e\) and \(x\). The first term is \(e\)
DLS
  • DLS
Taylor series gives 0?
experimentX
  • experimentX
gives -e/2 if you calculate correctly
ParthKohli
  • ParthKohli
Oh wait! lol, yeah. I forgot to consider that -e/2 which had no \(x\)
DLS
  • DLS
How about a different way rather than taylor series?
ParthKohli
  • ParthKohli
Definition of derivative? :-|
DLS
  • DLS
o.O
ParthKohli
  • ParthKohli
I was just thinking how this might be the definition of a derivative in disguise.
experimentX
  • experimentX
Binomial expansion might work ... but that's too long
ParthKohli
  • ParthKohli
DLS Use whatever you want. 15 minutes ago DLS How about a different way rather than taylor series? 9 minutes ago
DLS
  • DLS
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 :)
experimentX
  • experimentX
L'hopital works
experimentX
  • experimentX
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

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