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 September 8th, 2019, 06:45 AM #1 Newbie   Joined: Sep 2018 From: Spain Posts: 17 Thanks: 0 Natural Logarithms with limits Hi, I am a bit confused about how to solve this equation below Limit x-->2 ln(x-1) _______ x-2 Last edited by skipjack; September 8th, 2019 at 11:17 PM.
 September 8th, 2019, 07:31 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,838 Thanks: 653 Math Focus: Yet to find out.
 September 8th, 2019, 09:11 AM #3 Newbie   Joined: Sep 2018 From: Spain Posts: 17 Thanks: 0 I need to see the working. Last edited by skipjack; September 8th, 2019 at 11:18 PM.
 September 8th, 2019, 09:32 AM #4 Senior Member   Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 You can rewrite it as $\displaystyle \lim_{y\to 0} \frac{\ln(1+y)}{y}$ (by no means necessary, I just find the expression simpler that way). I would then use l'Hôpital's rule. $\displaystyle \lim_{y\to 0} \frac{\ln(1+y)}{y} = \lim_{y\to 0} \frac{\frac{d}{dy}\ln(1+y)}{\frac{dy}{dy}} = \lim_{y\to 0}\frac{\frac{1}{1+y}}{1} = 1$ Last edited by skipjack; September 8th, 2019 at 11:22 PM.
 September 8th, 2019, 12:32 PM #5 Newbie   Joined: Sep 2018 From: Spain Posts: 17 Thanks: 0 limits Wow, that was great! Thanks man. The l'Hôpital's rule, isn't that learnt in derivatives? I’ve never seen that method before. I just started limits. Last edited by skipjack; September 8th, 2019 at 11:24 PM.
 September 8th, 2019, 04:53 PM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond $\displaystyle \log\left[\lim_{y\to0}\left((1+y)^{1/y}\right)\right]=\log e=1$
September 8th, 2019, 05:20 PM   #7
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Quote:
 Originally Posted by DarnItJimImAnEngineer $\displaystyle \lim_{y\to 0} \frac{\ln(1+y)}{y} = lim_{y\to 0} \frac{\frac{d}{dy}\ln(1+y)}{\frac{dy}{dy}} = \lim_{y\to 0}\frac{\frac{1}{1+y}}{1} = 1$

$$\lim_{y\to 0} \frac{ln(1+y)}{y} = \left.\frac{\mathrm d}{\mathrm dx}\log{(1+x)}\right|_{x=0}$$
So your method is somewhat circular. You may as well just write down the result.

Last edited by skipjack; September 8th, 2019 at 11:29 PM.

 September 8th, 2019, 06:46 PM #8 Senior Member   Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 Greg, I'm guessing that's a well-known identity? I guess it doesn't require calculus, but I'd never come up with it in a million years. Archie, it looks like your second term and mine are the same (if you "cancel out" the /dy bits). I know historically, I'm going backwards, but we don't need to prove l'Hôpital's rule here; I know it works, so I'm using it. And raven, yes, it doesn't make much sense to use l'Hôpital if you haven't learned derivatives yet.
 September 8th, 2019, 07:55 PM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra It's not about proving l'Hôpital.
September 8th, 2019, 09:09 PM   #10
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Quote:
 Originally Posted by DarnItJimImAnEngineer Greg, I'm guessing that's a well-known identity?
Yes, it is. I've often heard of it referred to as "the limit definition of $e$".

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