Landau's notation for truncated taylor expansions I cannot seem to understand the difference between $f(h)=O(h)$ and $f(h)=o(h)$. The context in which I am learning about Landau's notation is in truncated Taylor expansions. Any input would be appreciated!
 A: If $f(x)$, $g(x)$, and $h(x)>0$ are all defined in a punctured neighborhood of the point $\xi$ ($\xi=\infty$ allowed) one writes
$$f(x)=g(x)+o\bigl(h(x)\bigr)\qquad(x\to\xi)\tag{1}$$
iff
$$\lim_{x\to\xi}{\bigl|f(x)-g(x)\bigr|\over h(x)}=0\ .\tag{2}$$
This means that $(1)$ is a denominator free typographical picture of the precise fact $(2)$. It expresses in an intuitive way that as $x\to\xi$ the difference $\bigl|f(x)-g(x)\bigr|$ becomes ever more negligible compared with  $h(x)$'s  order of magnitude.
In general formulas of type $(1)$ are used in the following circumstances: We are studying, or try to understand, a "complicated" or "general" function $f(x)$, and compare it with a "well understood" function $g(x)$, e.g., $x\mapsto e^x$, or the first few terms of the Taylor expansion of $f$. The function $h(x)$ that appears in $(1)$ is usually a "very simple" function that describes a typical asymptotic behavior, like linear, exponential, or logarithmic.
Looking at the Taylor expansion of an $f$ which is sufficiently differentiable at the origin we then can say that for every $n\geq0$ one has
$$f(x)=j_0^nf(x)+o(|x|^n)\qquad(x\to0)\ ,$$
where $$j_0^nf(x):=\sum_{k=0}^n{f^{(k)}(0)\over k!}\>x^k$$
is the $n^{\rm th}$ Taylor polynomial of $f$ at $0$.
Similarly: If $f(x)$, $g(x)$, and $h(x)>0$ are all defined near $\xi$ ($\xi=\infty$ allowed) one writes
$$f(x)=g(x)+O\bigl(h(x)\bigr)\qquad(x\to\xi)\tag{3}$$
iff there is some constant $C$ such that
$${\bigl|f(x)-g(x)\bigr|\over h(x)}\leq C\qquad(x\in\dot U)\tag{4}$$
in some punctured neighborhood $\dot U$ of $\xi$. This means that $(3)$ is a denominator free typographical picture of the precise fact $(4)$. It expresses in an intuitive way that as $x\to\xi$ the difference $\bigl|f(x)-g(x)\bigr|$ has at most the order of magnitude of  $h(x)$.
A: *

*$f(x)=O(h(x))$ as $x\to \infty$ means that for some $k>0$ we have $|f(x)\leq k|h(x)|$ for all sufficiently large $x$ in the domain of $f$. For example \, $f(x)=O(1)$ as $x\to \infty$ means  $\exists k>0\;\exists r\;\forall x>r\;(|f(x)|\leq k).$ The idea is that even if $|f(x)|\to \infty$ as $x\to \infty$  it does not grow faster than some fixed multiple of $|h(x)|$

*$f(x)=o(h(x)) $ as $x\to \infty$ means that for all $k>0$ we have $|f(x)|\leq k|h(x)|$ for all sufficiently large $x$. For example $f(x)=o(1)$ as $x\to \infty$ iff $ \lim_{x\to \infty} f(x)=0.$ If $h(x)\ne 0$ for all sufficiently large $x$, then $f(x)=o(h(x))$as $x\to \infty$ iff $\lim_{x\to \infty} f(x)/h(x)=0.$ The idea is that if $|f(x)|\to \infty$ as $x\to \infty$ then $|h(x)|$ goes to $\infty$ much faster, in the sense that $\frac{f(x)}{h(x)}\to 0.$

*For real number $ A$ : $f(x)=O(h(x))$ as $x\to A$ means that for some $k>0$ and some $d>0$ we have $[x\in Dom (f)\land0<|x-A|]\implies |f(x)|\leq k|h(x)|.$ In some contexts we may allow $x=A$ in the last sentence. In some contexts we may restrict $x$ to values greater than $A.$ 
Examples. $x-\sin x=O(x^3)$  and $1+\frac{1}{x}= O\left(\frac{1}{x}\right)$ as $x\to 0.$

*For real $A$ : $f(x)=o(h(x))$ as $x\to A$ means that for every $k>0$ there exists $d>0$ such that $[x\in Dom (f)\land 0<|x-A|<d]\implies |f(x)\leq k|h(x)|.$ As before there may be contextual restrictions on $x$.
Examples. $\log |x|=o\left(\frac{1}{x}\right)$ and $x-\sin x=o(x^2)$ as $x\to 0.$ 
