Limits with the natural logarithm

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SUMMARY

The discussion focuses on calculating limits involving the natural logarithm, specifically using the known limit \(\lim_{x\rightarrow 0}\frac{\ln(1+x)}{x}=1\) to derive three additional limits. The limits in question are \(\lim_{x\rightarrow 0}\frac{\ln(1-x)}{x}\), \(\lim_{x\rightarrow 0}\frac{\ln(1+x^{2})}{x}\), and \(\lim_{x\rightarrow 0}\frac{\ln(1+2x)}{x}\). Participants suggest using change of variables and substitutions to simplify the calculations, emphasizing the importance of recognizing patterns in limit problems.

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Yankel
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Hello

I have three limits to calculate, based on a given limits. What I know is:

\[\lim_{x\rightarrow 0}\frac{ln(1+x)}{x}=1\]

And based on this, I need to find (without L'Hopital rule), the following:

\[\lim_{x\rightarrow 0}\frac{ln(1-x)}{x}\]

\[\lim_{x\rightarrow 0}\frac{ln(1+x^{2})}{x}\]

\[\lim_{x\rightarrow 0}\frac{ln(1+2x)}{x}\]

I can't figure out the technique of moving from the known limits to the ones I need to find.

Thank you in advance !
 
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Hi Yankel,

you can try using change of variables for all of them, i.e $u=-x$ for the first one, $u=x^2$ for the second one, etc.
 
Ok, so we're given $\lim_{x\to0}\frac{\log(1+x)}{x}=1$ and we have

$$\lim_{x\to0}\frac{\log(1-x)}{x}=\lim_{x\to0}\frac{\log(1+(-x))}{x}=-\lim_{x\to0}\frac{\log(1+(-x))}{-x}$$

Can you compute it now? Can you make progress on the others, both with this method and by using substitution as Rido12 suggested?
 
So simple once you see an example...

Thank you !
 

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