Natural Logarithms with limits

Sep 2018
19
0
Spain
Hi, I am a bit confused about how to solve this equation below

Limit
x-->2 ln(x-1)
_______
x-2
 
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Sep 2018
19
0
Spain
I need to see the working.
 
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Jun 2019
493
262
USA
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.
\(\displaystyle \lim_{y\to 0} \frac{\ln(1+y)}{y} = \lim_{y\to 0} \frac{\frac{d}{dy}\ln(1+y)}{\frac{dy}{dy}} = \lim_{y\to 0}\frac{\frac{1}{1+y}}{1} = 1\)
 
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Sep 2018
19
0
Spain
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.
 
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greg1313

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Oct 2008
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\(\displaystyle \log\left[\lim_{y\to0}\left((1+y)^{1/y}\right)\right]=\log e=1\)
 

v8archie

Math Team
Dec 2013
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Colombia
\(\displaystyle \lim_{y\to 0} \frac{\ln(1+y)}{y} = lim_{y\to 0} \frac{\frac{d}{dy}\ln(1+y)}{\frac{dy}{dy}} = \lim_{y\to 0}\frac{\frac{1}{1+y}}{1} = 1\)

$$\lim_{y\to 0} \frac{ln(1+y)}{y} = \left.\frac{\mathrm d}{\mathrm dx}\log{(1+x)}\right|_{x=0}$$
So your method is somewhat circular. You may as well just write down the result.
 
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Jun 2019
493
262
USA
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.
 

v8archie

Math Team
Dec 2013
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It's not about proving l'Hôpital.