baldymcgee6
  • baldymcgee6
Tricky limit?
Mathematics
chestercat
  • chestercat
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baldymcgee6
  • baldymcgee6
\[\lim_{x\to 0}\left(\frac{(1+x)^{\frac{1}{x}}}{e}\right)^{\frac{1}{x}}.\]
baldymcgee6
  • baldymcgee6
sorry, thats kind of hard to see.
baldymcgee6
  • baldymcgee6
\[\LARGE \lim_{x\to 0}\left(\frac{(1+x)^{\frac{1}{x}}}{e}\right)^{\frac{1}{x}}.\]

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anonymous
  • anonymous
\[ \Large e = \lim_{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n \]When you reparameterize: \(x=1/n\) \[\Large e = \lim_{x \rightarrow 0}\left(1+x\right)^\frac{1}{x} \]
anonymous
  • anonymous
So it's a matter of settling that outer 1/x
baldymcgee6
  • baldymcgee6
I haven't learned 'reparameterize' yet, so I would assume I wouldn't have to use that... Supposed to use L'Hospital's rule
anonymous
  • anonymous
Okay, then you need to have an indeterminate form of \(\infty /\infty\) or \(0/0\)
baldymcgee6
  • baldymcgee6
right.
anonymous
  • anonymous
|dw:1353045336700:dw|
anonymous
  • anonymous
|dw:1353045499277:dw|
baldymcgee6
  • baldymcgee6
@mahmit2012 I am supposed to use L'Hospital's rule.
anonymous
  • anonymous
I can't read that bottom line.
baldymcgee6
  • baldymcgee6
"and in this case it is 1/sqrt(e)
anonymous
  • anonymous
No, of the previous picture
baldymcgee6
  • baldymcgee6
I don't know
anonymous
  • anonymous
Okay, let's start with bringing in the \(1/x\) to the numerator and denominator and figuring out if that is an indeterminate form.
anonymous
  • anonymous
|dw:1353047217870:dw|
anonymous
  • anonymous
|dw:1353047297672:dw|
baldymcgee6
  • baldymcgee6
@waterineyes can you help me understand this maybe?

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