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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 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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    Cheers Gerry, appreciate your help!2012-02-23