I'm trying to find the limit of the ratio of two functions
$ \lim\_{t \rightarrow \infty} \frac{f(t)}{g(t)} $ but only have the initial conditions and the differential equations they solve, but the equations don't have easy closed form solutions:


$f'(t) = -\beta f(t)^2 + (\beta - \gamma) f(t) + \gamma e^{-\delta t} \\
g'(t) = -\beta g(t)^2 + (\beta - \gamma) g(t) $


$f(0)=1, g(0)=1$ and $f(t) \rightarrow 0$ and $g(t) \rightarrow 0$.


L'hopital's is no use, and trying to numerically solve this for long time by solving the top and bottom first is no good, it breaks down after a point. I've consider the method of dominant balance, but I'm not sure it is applicable. I don't know what else to try, I'm a statistician out of his depth.



Limiting Ratio of Solutions to Ordinary Differential Equations
Your differential equations do have closed-form solutions. $f(t)$ is a rather complicated expression involving Bessel functions, while
$$ g \left( t \right) ={\frac {{{\rm e}^{- \left( \gamma-\beta \right) t
}} \left( \gamma-\beta \right) }{{\gamma-{\rm e}^{- \left( \gamma-\beta
 \right) t}}\beta}}
$$


Since you want $g(t) \to 0$ (presumably as $t \to + \infty$), I suppose you want
$\gamma > \beta$. I'll also assume $\beta, \gamma, \delta, c > 0$. As $t \to \infty$ we have $$g(t) \sim 
e^{-(\gamma-\beta)t} (1 - \beta/\gamma)$$


Now suppose $f(t) = u(t) g(t)$. I get


$$ u'(t) = -{\frac { \left( u \left( t \right) \right) ^{2} \left( \gamma-\beta
 \right) \beta}{\gamma\,{{\rm e}^{ \left( \gamma-\beta \right) t}}-
\beta}}+{\frac {u \left( t \right) \left( \gamma-\beta \right) \beta
}{\gamma\,{{\rm e}^{ \left( \gamma-\beta \right) t}}-\beta}}-{\frac {c
 \left( 2\,{{\rm e}^{ \left( \gamma-\beta-\delta \right) t }}\gamma\,
\beta-{{\rm e}^{\left( 2\,\gamma-2\,\beta-\delta \right) t}}{\gamma}^
{2}-{{\rm e}^{-\delta\,t}}{\beta}^{2} \right) }{ \left( \gamma-\beta
 \right) \left( \gamma\,{{\rm e}^{ \left( \gamma-\beta \right) t}}-
\beta \right) }}
$$


For large $t$, the coefficients of the first two terms on the right go to $0$ exponentially, while the largest contribution to the third is 
$$ \dfrac{c \gamma}{\gamma-\beta} e^{(\gamma-\beta-\delta)t}$$


So if $\gamma - \beta - \delta > 0$, $u$ will grow exponentially, while
if $\gamma - \beta - \delta < 0$, $u'$ will decay exponentially and $u$ will approach a constant. 


One nice special case is $\beta = \gamma/2$, $\delta = \gamma$, where 
I get
$$ \lim\_{t \to \infty} \dfrac{f(t)}{g(t)} = \dfrac{4\,\sqrt {c\gamma}\cosh \left( {\sqrt{2c/\gamma}}
 \right) +4\,\sqrt {2}c\,\sinh \left( {\sqrt{2c/\gamma}} \right) 
}{\sqrt{2}\,\gamma\sinh \left( {\sqrt{2c/\gamma}} \right) +2\,\sqrt {c\gamma}\cosh \left( {\sqrt{2c/\gamma}} \right) 
}
$$


