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We know that

$$ \lim\limits_{x \to 0} \frac{\ln(1+x)}{x}=1$$

so

$$ \lim\limits_{x \to 0} \ln(1+x)=x$$

Can I conclude that

$$ \lim\limits_{x \to 0} \frac{\ln(1-x)}{x}=1$$ and

$$ \lim\limits_{x \to 0} \ln(1-x)=-x$$

Is it correct?

Thanks

  • 3
    What do you mean by $\lim_{x\to 0} \ln(1+x)=x$? – 2017-01-06
  • 0
    The variable $s$ on the right hand side is out of scope. – 2017-01-06
  • 0
    The fourth line has no sense. – 2017-01-06

1 Answers 1

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What you can say is that $\ln(1 + x) \sim x$ as $x\to 0$, so $\ln(1 -x) \sim -x$ as $x\to 0$. This is the proper notation.