Given a master equation
$$\frac{\mathrm{d}P(n,t)}{\mathrm{d}t} = \sum_{n \neq n'}T(n|n')P(n',t) - \sum_{n \neq n'}T(n'|n)P(n,t) \tag{1} $$
$$ T(n\pm1|n) = \left(1-\frac{n}{N}\right) \left( \frac{n}{N}\right) \tag{2}$$
Making the substitution $x = n/N$, and expanding in powers of $1/N$, the goal is to derive the following,
$$\frac{\mathrm{\partial}P(x,t)}{\mathrm{\partial}t} = \left( x-\frac{1}{N}\right) \left(1-x+\frac{1}{N}\right)\left(P(x,t) -\frac{1}{N}\frac{\mathrm{\partial}P(x,t)}{\mathrm{\partial}x} +\frac{1}{2N^2}\frac{\mathrm{\partial}^2P(x,t)}{\mathrm{\partial}x^2} \right) +\left( x+\frac{1}{N}\right) \left(1-x-\frac{1}{N}\right)\left(P(x,t) +\frac{1}{N}\frac{\mathrm{\partial}P(x,t)}{\mathrm{\partial}x} +\frac{1}{2N^2}\frac{\mathrm{\partial}^2P(x,t)}{\mathrm{\partial}x^2} \right) -2x(1-x)P(x,t) $$
Following(Ref.1), I end up with
$$\frac{\mathrm{\partial}P(xN,t)}{\mathrm{\partial}t} = \left( x-\frac{1}{N}\right) \left(1-x+\frac{1}{N}\right)\left(P(xN,t) -\frac{\mathrm{\partial}P(xN,t)}{\mathrm{\partial}(1/N)} +\frac{1}{2}\frac{\mathrm{\partial}^2P(xN,t)}{\mathrm{\partial}(1/N)^2} \right). +\left( x+\frac{1}{N}\right) \left(1-x-\frac{1}{N}\right)\left(P(xN,t) +\frac{\mathrm{\partial}P(xN,t)}{\mathrm{\partial}(1/N)} +\frac{1}{2}\frac{\mathrm{\partial}^2P(xN,t)}{\mathrm{\partial}(1/N)^2} \right) -2x(1-x)P(xN,t) $$
Which is close but not quite the expression I want. How does one get rid of $xN$ in the argument for the probability $P$