Order of logarithmic functions as $x \rightarrow 0$

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 $\Theta$? –  Aryabhata Feb 23 '12 at 0:08
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. –  Michael Hardy Feb 23 '12 at 0:10
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). –  Heijden Feb 23 '12 at 0:14
@Heijden: This follows from the definition of $o$ and $\mathcal{O}$. –  Aryabhata Feb 23 '12 at 0:42
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. –  Gerry Myerson Feb 23 '12 at 0:44

$\log(1+u)$ 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 $x\to0$.
$(1+2x)/x(1-2x) \to \infty$ as $x \to 0$, so you cannot use the fact that $\log(1+x)$ is of the order of $x$. –  Aryabhata Feb 23 '12 at 0:53
Oops. ${}{}{}{}$ –  Gerry Myerson Feb 23 '12 at 0:59