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Let $$D(s) + KN(s) = 0 \tag{1}$$where $D(s)$ and $N(s)$ are polynomials of $s \in \mathbb{C}$ such that $\text{Deg}(D) = n, \ \text{Deg}(N) = m$ and $n\ge m$. The root locus method tells us how the solutions of $(1)$ changes as we change parameter $K$ from $K=0$ to $K = \infty$.

I'm trying to understand these extreme cases. Let $K \to 0$ and we have $$D(s) = 0$$ so in this case, the set of the solutions is $A = \{s \in \mathbb{R} | D(s) = 0\}$. Now let $K \to \infty$, if we choose $s$ such that $N(s) \not = 0$ then the answer will be infinity. So we should choose $s$ such that $N(s) = 0$. In that case, if $(1)$ holds, we also have $D(s) =0$ which means $N(s)$ and $D(s)$ have the same factor but this isn't the result that should be obtained. Curiously, if we rewrite $(1)$ $$\frac{N(s)}{D(s)} = -\frac{1}{K} \tag{2}$$ and let $K\to \infty $, one possible case that $(2)$ holds is $N(s) = 0$ which gives us $m$ solutions and this doesn't require $D(s)$ have the same factor as $N(s)$! Why this happens? And why the first solution is wrong?

Example: Let $D(s) = s^2 - 4$ and $N(s) = s + 3$. So $(1)$ becomes $$s^2 - 4 + K(s+3) = 0$$If $K \to \infty $ and $s = -3$ then $9 - 4 = 0$ which is clearly wrong. On the other hand, rewriting the equation $$\frac{s+3}{s^2 - 4} = -\frac{1}{K}$$ If $K \to \infty $ and $s = -3$ then $\frac{0}{9-4} = 0$ which is true, of course.

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    $\begingroup$ These visuals may offer some insight, though not directly addressing the question. $\endgroup$ Jan 5 at 12:23
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    $\begingroup$ @OverLordGoldDragon Nice visualization! Thanks. $\endgroup$
    – S.H.W
    Jan 5 at 13:10
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The problem with your example is that $\infty\cdot 0$ isn't necessarily equal to zero. The only way to judge what is happening in the limit $K\to\infty$ is to divide the original equation by $K$:

$$\frac{D(s)}{K}+N(s)=0\tag{1}$$

Now it is obvious that for $K\to\infty$ the actual value of $D(s)$ is irrelevant, as long as it is finite. Consequently, the only necessary condition for $(1)$ to be true as $K$ becomes large is that $N(s)=0$.

This problem is a bit similar to the problem of determining the limit

$$\lim_{x\to\infty}\frac{x}{x+c}\tag{2}$$

The value of the limit $(2)$ is independent of the choice of the constant $c$, which becomes negligible compared to $x$. The same is true in the first equation of the question: the actual value of $D(s)$ becomes irrelevant compared to $KN(s)$.

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  • $\begingroup$ Thanks. You are right that $\infty\cdot 0$ isn't necessarily equal to zero but my mean was that first choose $s$ such that $N(s)$ becomes identically zero and then $K \to \infty$. In this way we are not faced with the problem of $\infty\cdot 0$. $\endgroup$
    – S.H.W
    Jan 6 at 22:25
  • $\begingroup$ @S.H.W: But that's not the way limits work. In the form $D(s)+KN(s)=0$ you can't directly deal with the limit $K\to\infty$, you have to divide by $K$ to see what's happening. $\endgroup$
    – Matt L.
    Jan 7 at 12:33
  • $\begingroup$ I think the problem is that we shouldn't evaluate the expression before taking the limit since we are looking for $s$ such that solves $D(s) + KN(s) = 0$ for different values of $K$. So $s$ is unknown and we should first take the limit and then see what values of $s$ solve the equation. Do you think that this is the problem? $\endgroup$
    – S.H.W
    Jan 7 at 15:29
  • $\begingroup$ @S.H.W: Yes, but the problem is that the limit is not finite, so you can't get much information from that limit. The only way is really to divide by $K$. $\endgroup$
    – Matt L.
    Jan 7 at 16:07

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