$\ln(x)$ and Big O notation I have tried to assert that $\ln(x)=O(x^0)$ a few times, but it seems fairly obvious that this statement should be false, and so I've been faced with some rightful speculation.
My reason is that
$$\ln(x)=\lim_{k\to0}\frac{x^k-1}{k}=O(x^0)?$$
But if not, then I ask

What is the smallest value of $k$ for $\ln(x)=O(x^k)?$  A proof would be nice, and possibly some cleanuping if my notation is used incorrectly.

 A: The first point is: when you use asymptotic notations, you need to specify with regard to what point the asymptotics is taken. Here, it looks like this is when $x\to 0^+$; note that this could have equally been $x\to\infty$, so specifying it is required.
Now, you have that for any fixed $\alpha > 0$,
$$
\ln x = O(x^{-\alpha})
$$
when $x\to 0^+$, since $\left\lvert x^\alpha \ln x\right\rvert\xrightarrow[x\to0^+]{} 0$. Yet, this does not "go to the limit": there is no minimum $\alpha\geq 0$ for which this holds.

Parenthesis: you have $\ln x  = x^{\frac{\ln\ln x}{\ln x}}$, so you could technically write $\ln x = \Theta(x^{\frac{\ln\ln x}{\ln x}})$... but note that this means "taking $\alpha = \alpha(x) = \frac{\ln\ln x}{\ln x}$."
A: For $r>0$ take $M>1$ such that $1/M<r.$ For $y>M$ we have $$0<\ln y=\int_1^y (1/z)\;dz= \int_1^M(1/z)\;dz+\int_M^y(1/z)\;dz=$$ $$=\ln M+\int_M^y(1/z)\;dz<\ln M+\int_M^y(1/M)\;dz=\ln M+(y-M)/M.$$  $$\text { So, }\quad 0<(\ln y)/y<(\ln M)/y+(1-M/y)/M.$$ Therefore  $$0\leq \sup_{x\geq y}\;(\ln x)/x\leq \sup_{x\geq y}(\;(\ln M)/x+(1-M/y)/M\;)=$$ $$=(\ln M)/y+(1-M/y)/M.$$ $$\text {Therefore, }\quad 0\leq \lim_{y\to \infty}\sup_{x\geq y}\;(\ln x)/x\leq$$ $$\leq \lim_{y\to \infty}\;(\ln M)/y+(1-M/y)/M=1/M<r.$$ Since $r$ can be as small as desired, we have therefore $\lim_{y\to \infty}(\ln y)/y=0.$
For positive $a,b$ we have $$(\ln y)^a/y^b=((\ln y)/y^{b/a})^a=(((a/b)\ln y^{b/a})/y^{b/a})^a.$$ Let $z=y^{b/a}.$ Then $z\to \infty$ as $y\to \infty$ and we have $$(\ln y)^b/y^a=((a/b)^a\cdot ((\ln z)/z))^a\to 0$$ as $y$ (and $z$) go to $\infty.$ 
In particular with $a=1$ and $b>0$ we have $\lim_{y\to \infty}(\ln y)/y^b=0.$ So as $y\to \infty$ we have $\ln y=O(y^b)$ for any $b>0. $
Actually we have $\ln y=o(y^b)$ as $y\to \infty$ for any $b>0$ because  $\lim_{y\to \infty}(\ln y)/y^b=0.$
For $0<y<1$and $y\to 0$ let $x=1/y.$ Then  $|\ln y|/y=x\ln x$ and use the above inequalities.
