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September 8th, 2019, 06:45 AM   #1
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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.
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September 8th, 2019, 07:31 AM   #2
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https://www.desmos.com/calculator/whl8oybhms
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September 8th, 2019, 09:11 AM   #3
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I need to see the working.

Last edited by skipjack; September 8th, 2019 at 11:18 PM.
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September 8th, 2019, 09:32 AM   #4
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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.
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September 8th, 2019, 12:32 PM   #5
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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.
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September 8th, 2019, 04:53 PM   #6
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$\displaystyle \log\left[\lim_{y\to0}\left((1+y)^{1/y}\right)\right]=\log e=1$
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September 8th, 2019, 05:20 PM   #7
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Quote:
Originally Posted by DarnItJimImAnEngineer View Post
$\displaystyle \lim_{y\to 0} \frac{\ln(1+y)}{y} = lim_{y\to 0} \frac{\frac{d}{dy}\ln(1+y)}{\frac{dy}{dy}} = \lim_{y\to 0}\frac{\frac{1}{1+y}}{1} = 1$

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

Last edited by skipjack; September 8th, 2019 at 11:29 PM.
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September 8th, 2019, 06:46 PM   #8
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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.
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September 8th, 2019, 07:55 PM   #9
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It's not about proving l'Hôpital.
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September 8th, 2019, 09:09 PM   #10
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Quote:
Originally Posted by DarnItJimImAnEngineer View Post
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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