I am left wondering about part of a proof involving limits. Here is the math included (which I wish had more details!)
$$ \lim_{R\rightarrow \infty }S(Re^{j \theta})=\lim_{R\rightarrow \infty }\frac{1}{1+L(Re^{j \theta})}=\frac{1}{1+\lim_{s\rightarrow \infty }L(s)R^{-n_{r}}e^{-n_{r} j \theta }} $$
where $ n_{r} $ is the relative degree of $L(s)$.
For context $S(s)$ is the sensitivity transfer function and $L(s)$ is the open loop transfer function, from Control Theory.
I can't figure out how they go from the second term to the third term. Any help would be greatly appreciated. Thanks
For additional context, here is a link to the paper containing the proof. Bottom, righthand side is the equation in question.
https://www.sciencedirect.com/science/article/pii/S2405896315025045
Okay, here is what I came up with today. My confidence in its correctness is somewhere around luke warm.
$$ \lim_{R\rightarrow\infty}{S\left(Re^{j\theta}\right)}=\lim_{R\rightarrow\infty}{\frac{1}{1+L\left(Re^{j\theta}\right)}}=\frac{1}{1+\lim_{s\rightarrow\infty}{L\left(s\right)}R^{-n_r}e^{-n_rj\theta}} $$
To understand what is going on when going from the second term to the third term in the equation above, consider the following general form of the open-loop transfer function. $$ L\left(Re^{j\theta}\right)=\frac{H\left(R^{n_H}e^{j\theta n_H}\right)}{G\left(R^{n_G}e^{j\theta n_G}\right)} $$ The Open-Loop Transfer Function, L(s), is ultimately a complex function, to some order of z. The above equation shows that the numerator is of order nH and the denominator is of order nG. Let us take a closer look at this specific term in the denominator of the original equation in question. $$ \lim_{R\rightarrow\infty}{L\left(Re^{j\theta}\right)}=\lim_{R\rightarrow\infty}{\frac{H\left(R^{n_H}e^{j\theta n_H}\right)}{G\left(R^{n_G}e^{j\theta n_G}\right)}} $$ For the general transfer function, the numerator and denominator will be in terms of powers of z. Next, consider that in the limit of R going to infinity, it is ultimately the highest order power of R in the numerator and denominator, along with the coefficient associated with that term, which determines the value of the limit. We can express that term as the following. $$ \lim_{R\rightarrow\infty}{\frac{H\left(R^{n_H}e^{j\theta n_H}\right)}{G\left(R^{n_G}e^{j\theta n_G}\right)}}=\frac{A\ast R^{n_H}e^{j\theta n_H}}{B\ast R^{n_G}e^{j\theta n_G}}=\frac{A}{B}R^{\left(n_H-n_G\right)e^{j\theta\left(n_H-n_G\right)}} $$ There are two things to note about that last term. We are taking the limit of L(z) in the complex plane and arriving at the above result. Well, what if we also took the limit of L(s) in the s-plane as s goes to infinity. If you work through the same reasoning we used in the complex plane, you will eventually come to an answer of A/B! The other thing to note is that (nH - nG) is just the relative degree of L(z), or nr. That let’s us write all this up as: $$ \lim_{R\rightarrow\infty}{S\left(Re^{j\theta}\right)}=\frac{1}{1+L\left(\infty\right)\ast R^{-n_r}e^{-n_rj\theta}} $$ In the above, $$ L\left(\infty\right)=\lim_{s\rightarrow\infty}{L\left(s\right)}. $$