Taylor expansion of $H = \sqrt{m^2 - \hbar^2 \nabla^2}$ $$ H = \sqrt{m^2 - \hbar^2 \nabla^2} $$
Suppose that there is a equation like this. How do you taylor-expand this equation? I am extremely confused.

Use the expansion of $\sqrt{1-x}$ about $x=0$, 
$$\begin{eqnarray*}
H &=& \sqrt{m^2 - \hbar^2 \nabla^2} \\
&=& m\sqrt{1-\frac{\hbar^2\nabla^2}{m^2}} \\
&=& m\left(1-\frac{1}{2}\frac{\hbar^2\nabla^2}{m^2} 
 - \frac{1}{8} \frac{\hbar^4\nabla^4}{m^4}
 + \ldots\right) \\
&=& m
 -\frac{\hbar^2\nabla^2}{2m} 
 - \frac{\hbar^4\nabla^4}{8m^3}
 - \ldots
\end{eqnarray*}$$
If we put back the factors of $c$ we'll find this is an expansion in small 
$\hbar \nabla/(m c)$, i.e., in large $c$. 
Keeping the first two terms we recover Schrödinger's Hamiltonian for a free particle (up to an additional, unphysical constant).
