This is from a mathematical biology exam:
I’ve answered parts $a)$ to $f)$ absolutely fine but I’m struggling with what I presume is the final mark of part $f)$.
I’ve solved the equation directly to obtain:
$$N(t)=\frac{(1-k)N_0 e^{(1-k)t}}{1-k-N_0 +N_0e^{(1-k)t}}$$
Now I know I need to examine the cases when $1-k<0$ and $1-k>0$. I know that when $1-k<0$ clearly $N(t)\rightarrow 0$ but I don’t understand the behaviour when $1-k>0$. I know that $N(t)\rightarrow 1-k$ when $1-k>0$ as it needs to agree with the steady state but I just don’t understand how to show this. Could someone explain this please?
Dividing numerator and denominator of the right hand side of the equation $$N(t)=\frac{(1-k)N_0 e^{(1-k)t}}{1-k-N_0 +N_0e^{(1-k)t}}$$ by $N_0e^{(1-k)t}$ you get $$N(t)=\frac{(1-k)}{((1-k)/N_0-1)e^{-(1-k)t} +1}$$
As $1-k$ is supposed to be strictly positive, you have $$\lim\limits_{t \to \infty} ((1-k)/N_0-1)e^{-(1-k)t} = 0$$
Hence the desired result $\lim\limits_{t \to \infty} N(t) = 1-k$