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Callisto Group Title

Find the limit:\[\lim_{x \rightarrow \infty} (\frac{x+2}{x-1})^x\]

  • one year ago
  • one year ago

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  1. Callisto Group Title
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    Let \(y= (\frac{x+2}{x-1})^x\) \[ln y = x\ln(\frac{x+2}{x-1})\]Can't use L'Hopital's Rule now..Hmm..

    • one year ago
  2. AccessDenied Group Title
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    I was thinking something like this: \( \displaystyle \frac{x + 2}{x - 1} = \frac{x - 1 + 3}{x - 1} = \frac{x - 1}{x - 1} + \frac{3}{x - 1} = 1 + \frac{3}{x-1}\) The end expression looks a lot more like the definition of e in terms of limits (lim as x appraoches infinity of (1 + 1/x)^x

    • one year ago
  3. zepdrix Group Title
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    You CAN apply L'Hop Rule Callisto, we just need to apply a tiny trick from the spot you got to. You want to write the x as the reciprocal of the reciprocal of x.\[\huge x\ln\left(\frac{x+2}{x-1}\right)=\frac{\ln\left(\frac{x+2}{x-1}\right)}{\frac{1}{x}}\] And you'll notice that from here, we get an indeterminate form, 0/0 which allows us to apply L'Hop.

    • one year ago
  4. Callisto Group Title
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    @zepdrix My bad, I should have revised more... Thanks a ton!!! @AccessDenied I haven't thought of this way... :'( Perhaps I should give a try!!

    • one year ago
  5. zepdrix Group Title
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    Definition of e? :) Hmm that's clever, yah I like that.

    • one year ago
  6. AccessDenied Group Title
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    Yea, I vaguely recall seeing it done somewhere, it seemed like it would be pretty useful here. :P

    • one year ago
  7. zepdrix Group Title
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    I remember it comes up if you need to use the Limit Definition of a Derivative to show the derivative of ln x. That one is kinda fun :3

    • one year ago
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