o(h) term in birth-death process This is from the note

I have two questions.


*

*Since $o(h)$ represents the probability of 2 birth and 1 death, 3 birth and 2 death, etc, why it still says $P(|X(t+h)-X(t)|>1)=o(h)$? shouldn't it be $P(|X(t+h)-X(t)|=1)=o(h)$?

*how to get the equation 13 from equation 12?
 A: First of all, what does $o(h)$ (it is a class of functions, called Landau or Big- O notation) actually mean? Generally it means for a function $f\in o(g)$ that we have
$$
{\displaystyle \lim _{x\to a}\left|{\frac {f(x)}{g(x)}}\right|=0}
$$ As an example, take a differentiable function $f$, then we have for $h\to0$ the following 
$$
f(x+h)=f(x)+hf'(x)+o(h)
$$
this means, that the approximation error goes faster than linear to $0$.
Your $2$nd question was already answered by @Math1000, for completeness:
Since we investigate the behavior of $h$ close to $0$, we indeed have 
$$
h^2\ll h \text{ respectively for }n\geq2: h^n\ll h
$$ 
so for $h\to0$, all higher powers $h^n$ are of neglectable magnitude compared to $h$ and therefore we get 
$$
(\lambda_ih)(1-h(\lambda_i+\mu_i)+\ldots)+o(h)=\lambda_ih+o(h)
$$
Your $1$st question:
The expression
$$
P(|X(t+h)-X(t)|>1|X(t)=i)=o(h)
$$
means, that for sufficient small h  ($\to 0$), the probability that there occurs  a jump in each direction larger of magnitude larger than $1$ is converging to $0$ faster than linear and for very small $h$ neglectable.
The bottom line is, that a Birth-Death process only jumps by $\pm1$ into the next state. 
