Here's the question you clicked on:
Callisto
Find the limit:\[\lim_{x \rightarrow \infty} (\frac{x+2}{x-1})^x\]
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..
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
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.
@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!!
Definition of e? :) Hmm that's clever, yah I like that.
Yea, I vaguely recall seeing it done somewhere, it seemed like it would be pretty useful here. :P
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