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anonymous
 3 years ago
Tricky limit?
anonymous
 3 years ago
Tricky limit?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x\to 0}\left(\frac{(1+x)^{\frac{1}{x}}}{e}\right)^{\frac{1}{x}}.\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry, thats kind of hard to see.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\LARGE \lim_{x\to 0}\left(\frac{(1+x)^{\frac{1}{x}}}{e}\right)^{\frac{1}{x}}.\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \Large e = \lim_{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n \]When you reparameterize: \(x=1/n\) \[\Large e = \lim_{x \rightarrow 0}\left(1+x\right)^\frac{1}{x} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So it's a matter of settling that outer 1/x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I haven't learned 'reparameterize' yet, so I would assume I wouldn't have to use that... Supposed to use L'Hospital's rule

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, then you need to have an indeterminate form of \(\infty /\infty\) or \(0/0\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353045336700:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353045499277:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@mahmit2012 I am supposed to use L'Hospital's rule.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I can't read that bottom line.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"and in this case it is 1/sqrt(e)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, of the previous picture

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, let's start with bringing in the \(1/x\) to the numerator and denominator and figuring out if that is an indeterminate form.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353047217870:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353047297672:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@waterineyes can you help me understand this maybe?
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