How would I compute the limit of a function such as this? $$\lim_{m\to \infty}\frac{m}{((x+m)(g-1)-m(g-2))(1+x+m)-m^2}$$
I'm not sure what to divide by in this case as I would just get the limit of the numerator to be either $0$ or $\infty$. The solution should be :
$$\lim_{m\to \infty} f(m) = \frac{1}{1+gx}$$
I'm thinking to get to this, $f(m)$ should be simplified further.
On
Compute the derivative of the denominator, call it $F(m)$: \begin{align} F(m) &=(g-1-g+2)(1+x+m)+(x+m)(g-1)-m(g-2)-2m\\ &=1+x+x(g-1)+m(1+g-1-g+2-2)\\ &=1+xg \end{align} Now it's just l'Hôpital. Or, if you prefer, $F(m)=(1+xg)m+A$, for some constant term $A$, precisely $$ A=F(0)=x(g-1)(1+x) $$ but it's unimportant.
Note that
$$\lim_{m\to \infty}\frac{m}{g m x + g x^2 + g x + m - x^2 - x}=\\=\lim_{m\to \infty}\frac{1}{g x + (g x^2)/m + (g x)/m + 1 - x^2/m - x/m}=\frac1{gx+1}$$