Limits of a parametric equation I have the following equation which states that $σ$ is an explicit function of $k$:$$σ=\frac b{1-c\left(1+\left(\fracβαk^A-B\right)^{-1}\right)},$$
where $A=\dfrac{b+c-1}{b}, \;B=\dfrac{c}{b+c-1},\;\;\dfracβα$ is a constant and the value of $σ$ has to be positive.
The author found that:


*

*If $b+c>1$, the value of $\sigma$ declines with increased $k$ and approaches $\dfrac{b}{1-c}$ which is greater than $1$ as $k$ increases without limit. 

*If $b+c<1$, the value of $\sigma$ increases from $\dfrac{b}{1-c}$ which is less than $1$, to $1$ as $k$ increases from $0$ to infinity.


How can I derive these results mathematically?
Further I have tried to simulate the function. While I can replicate the result when $b+c<1$ with the function that tends to the limit $\dfrac{b}{1-c}$, I see strange behaviors not corresponding to the result described above when $b+c>1$. So, are there any restrictions so that the values of $b$ and $c$ whose sum is greater than one produce a decreasing function and tending to the $\dfrac{b}{1-c}$ limit which this time should be greater than $1$?
Thanks a lot in advance to anyone who wants to help me :)
 A: It is not given, though I assume $b,c>0.$ 


*

*If $\;b+c>1\;$ then $A=\frac{b+c-1}{b}>0,$ and consequently $k^A\;$ increases, $\;k^A \to \infty\;$ as $\;k\to \infty.$
Then $\left(\fracβαk^A-B\right)^{-1}\;$ decreases to $0$ and finally $\sigma\;$ decreases to $$\lim_{k\to\infty}\frac b{1-c\left(1+\left(\fracβαk^A-B\right)^{-1}\right)}=\frac{b}{1-c\cdot (1+0)}=\frac{b}{1-c}$$

*If $\;b+c<1\;$ then $A=\frac{b+c-1}{b}<0,$ and $k^A\;$ decreases to $0$ as $\;k\to \infty.$
Note that  $\lim\limits_{k\to 0^{+}}k^A=\infty.$
Further, $\left(\fracβαk^A-B\right)=\left(\fracβαk^A+\frac{c}{1-(b+c)}\right)\;$ decreases to $\frac{c}{1-(b+c)}\;$ from where we conclude that $\left(\fracβαk^A-B\right)^{-1}\;$ increases to $\;\frac{1-(b+c)}{c}.$
Put together, $\sigma$ increases from the value$$\lim_{k\to 0^{+}}\sigma(k)=\lim_{k\to 0^{+}}\frac b{1-c\left(1+\left(\fracβαk^A-B\right)^{-1}\right)}=\frac{b}{1-c\left(1+0\right)}=\frac{b}{1-c}$$ 
to
$$\lim_{k\to\infty}\sigma(k)=\lim_{k\to\infty}\frac b{1-c\left(1+\left(\fracβαk^A-B\right)^{-1}\right)}=\frac{b}{1-c\left(1+\frac{1-(b+c)}{c}\right)}=1.$$ 
