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Callisto
 2 years ago
Best ResponseYou've already chosen the best response.0Let \(y= (\frac{x+2}{x1})^x\) \[ln y = x\ln(\frac{x+2}{x1})\]Can't use L'Hopital's Rule now..Hmm..

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.2I was thinking something like this: \( \displaystyle \frac{x + 2}{x  1} = \frac{x  1 + 3}{x  1} = \frac{x  1}{x  1} + \frac{3}{x  1} = 1 + \frac{3}{x1}\) 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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2You 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.\[\huge x\ln\left(\frac{x+2}{x1}\right)=\frac{\ln\left(\frac{x+2}{x1}\right)}{\frac{1}{x}}\] And you'll notice that from here, we get an indeterminate form, 0/0 which allows us to apply L'Hop.

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.0@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!!

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2Definition of e? :) Hmm that's clever, yah I like that.

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.2Yea, I vaguely recall seeing it done somewhere, it seemed like it would be pretty useful here. :P

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2I 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
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