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baldymcgee6

  • 3 years ago

Tricky limit?

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  1. baldymcgee6
    • 3 years ago
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    \[\lim_{x\to 0}\left(\frac{(1+x)^{\frac{1}{x}}}{e}\right)^{\frac{1}{x}}.\]

  2. baldymcgee6
    • 3 years ago
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    sorry, thats kind of hard to see.

  3. baldymcgee6
    • 3 years ago
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    \[\LARGE \lim_{x\to 0}\left(\frac{(1+x)^{\frac{1}{x}}}{e}\right)^{\frac{1}{x}}.\]

  4. wio
    • 3 years ago
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    \[ \Large e = \lim_{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n \]When you reparameterize: \(x=1/n\) \[\Large e = \lim_{x \rightarrow 0}\left(1+x\right)^\frac{1}{x} \]

  5. wio
    • 3 years ago
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    So it's a matter of settling that outer 1/x

  6. baldymcgee6
    • 3 years ago
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    I haven't learned 'reparameterize' yet, so I would assume I wouldn't have to use that... Supposed to use L'Hospital's rule

  7. wio
    • 3 years ago
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    Okay, then you need to have an indeterminate form of \(\infty /\infty\) or \(0/0\)

  8. baldymcgee6
    • 3 years ago
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    right.

  9. mahmit2012
    • 3 years ago
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    |dw:1353045336700:dw|

  10. mahmit2012
    • 3 years ago
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    |dw:1353045499277:dw|

  11. baldymcgee6
    • 3 years ago
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    @mahmit2012 I am supposed to use L'Hospital's rule.

  12. wio
    • 3 years ago
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    I can't read that bottom line.

  13. baldymcgee6
    • 3 years ago
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    "and in this case it is 1/sqrt(e)

  14. wio
    • 3 years ago
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    No, of the previous picture

  15. baldymcgee6
    • 3 years ago
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    I don't know

  16. wio
    • 3 years ago
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    Okay, let's start with bringing in the \(1/x\) to the numerator and denominator and figuring out if that is an indeterminate form.

  17. mahmit2012
    • 3 years ago
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    |dw:1353047217870:dw|

  18. mahmit2012
    • 3 years ago
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    |dw:1353047297672:dw|

  19. baldymcgee6
    • 3 years ago
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    @waterineyes can you help me understand this maybe?

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