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I am trying to find the order of logarithmic expressions as $x \rightarrow 0$. For example I can find that $\ln(1+x) = \mathcal{O}(x)$ and $\ln(1+x) = \mathcal{o}(1)$.

But when dealing with more complicated expressions below:

$$\ln\left[ 1+\frac{\ln(1+2x)}{1-2x} \right]$$ and $$\ln\left[ 1+\frac{1+2x}{x(1-2x)} \right].$$

I can find $\ln\left[ 1+\frac{\ln(1+2x)}{1-2x} \right]=\mathcal{o}(1)$, but how would I determine $\mathcal{O}$, and also for the 2nd expression?

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    If $f(x) = o(1)$, then $f(x) = \mathcal{O}(1)$ too. Are you looking for $\Theta$?2012-02-23
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    I changed "ln" to "\ln" throughout. Notice this difference: $w ln x$ and $w \ln x$. The former is coded as "w ln x" and the latter as "w \ln x". The backslash has two effects (1) "ln" does not get italicized, and (2) proper spacing precedes and follows it.2012-02-23
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    Michael Hardy Thanks. @Aryabhata Thanks. No Im not looking for Big Theta, just Big O and Little O. May I ask how you got that $f(x) = \mathcal{O}(1)$? Because as $x \rightarrow 0$, the first expression goes to 0 (not bounded).2012-02-23
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    @Heijden: This follows from the definition of $o$ and $\mathcal{O}$.2012-02-23
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    If it goes to zero, then it certainly is bounded above, which is exactly what big-oh of 1 does. Have another look at the definitions of big-oh and of theta.2012-02-23

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$\log(1+u)$ is on the order of $u$ as $u\to0$, so your first problem is on the order of $(\log(1+2x))/(1-2x)$, which is on the order of $x/(1-2x)$, which is on the order of $x$.

The second one is on the order of $(1+2x)/(x(1-2x))$, which is on the order of $1/x$.

EDIT: as noted in the comments, the second exposition is wrong. The argument of the logarithm is on the order of $1/x$ as $x\to0$, so I think all you can say is the expression is on the order of $-\log x$ as $x\to0$.

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    $(1+2x)/x(1-2x) \to \infty$ as $x \to 0$, so you cannot use the fact that $\log(1+x)$ is of the order of $x$.2012-02-23
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    Oops. ${}{}{}{}$2012-02-23
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    Cheers Gerry, appreciate your help!2012-02-23