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(x1) _______ x2 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  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra  Quote:
$$\lim_{y\to 0} \frac{ln(1+y)}{y} = \left.\frac{\mathrm d}{\mathrm dx}\log{(1+x)}\right_{x=0}$$ 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 wellknown 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 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond  

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limits, logarithms, natural 
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