[m] (\frac{2x+2z}{x+z+3}-\frac{2x}{x+3})=\frac{(2x+2z)(x+3)-2x(x+z+3)}{(x+z+3)(x+3)}=\frac{2x^2+2xz+6x+6z-2x^2-2xz-6x}{(x+z+3)(x+3)}=\frac{6z}{(x+z+3)(x+3)}[/m]
[m]lim_{z → 0}(\frac{2x+2z}{x+z+3}-\frac{2x}{x+3}):z=lim_{z → 0}\frac{6z}{z(x+z+3)(x+3)}=\frac{6}{(x+3)^2}[/m]