Solutions of $f(k,s)=0$ as $k\to\infty$ Let $f(k,s) = D(s) + kN(s)$ where $D(s)$ and $N(s)$ are polynomials of $s \in \mathbb{C}$ such that $\text{Deg}(D) = n, \ \text{Deg}(N) = m$ and $n\ge m$. Also $D(s)$ and $N(s)$ doesn't have common factor. Find the values of $s$ such that $$\lim_{k\to \infty}f(k,s) = 0  \tag{1}$$ My try: Assume that $N(s) \not = 0$. Obviously, in this case $\lim_{k\to \infty}f(k,s)$ is divergent for any $s$. Now suppose $N(s) = 0$ and we have $\lim_{k\to \infty}f(k,s) = D(s)$. So the answer is $D(s) = N(s) = 0$ which is impossible since $D(s)$ and $N(s)$ doesn't have common factor. What's wrong in my reasoning?
Also is $(1)$ equivalent to find $s$ such that $f(k,s) = 0$ and then $\lim_{k \to \infty} s(k)$? I mean that we first find the roots of $f(k,s)$ with respect to $k$ and see what happens to the roots when $k \to \infty$.
Edit: The correct answer is as follows. Divide the equation $D(s) + kN(s) = 0$ by $k$ and let $k \to \infty$. So we have $$\lim_{k \to \infty}(\frac{D(s)}{k} + N(s)) = N(s) = 0$$ I don't know why my argument is flawed and whether the mentioned equivalence holds.
Edit2: Thanks to @Alex Ravsky, it turns out that the true question is

What are the solutions of $f(k,s)=0$ as $k\to\infty$?

 A: If $N$ and $D$ are relatively prime, then $q(s)N(s)+r(s)D(s)=1$ for some $q(s),r(s)\in \mathbb{C}[s]$.
Using this, $r(s)f(k,s)$ can be re–written as $r(s)f(k,s)=1+(kr(s)-q(s))N(s)$. Now if $N(s)\not =0$, then $\lim_{k\rightarrow \infty} q(s)f(k,s)\not =0$ because $kr(s)-q(s)\not =0$ by taking $k$ large enough for a fixed $s$. Furthermore, as $k\rightarrow \infty$ for any fixed value the RHS blows up.

After writing this, I am wondering why you are interested in this question since it numerical experimentation should indicate that the polynomial blows up.
A: 
What are the solutions of $f(k,s)=0$ as $k\to\infty$?

It can happen that the only root $s(k)$ of $f(k,s)$ is $k$ (for instance, when $D(s)=s$ and $N(s)=-1$), then  $\lim_{k \to \infty} s(k)=\infty$.
But in order to develop the argument from the proposed answer we suppose that there exists $R\ge 0$ such that for each $k$ the equation $f(k,s)=0$  has a root $s(k)$ contained in a set $K=\{s\in\Bbb C:|z|\le R\}$. Then for any $k\ne 0$ we have $\frac{D(s(k))}{k} + N(s(k))=0$. Since the set $K$ is compact and a function $D$ is continuous, there exists $M>0$ such that $|f(s)|\le M$ for each $s\in K$. This follows $\lim_{k \to \infty} N(s(k))=0$. Let $S=\{s\in K: N(s)=0\}$.
We can describe an asymptotic behavior of $s(k)$ as follows. For each $\varepsilon>0$ there exists natural $k_\varepsilon$ such that for any $k> k_\varepsilon$ there exists $s\in S$ with $|s(k)-s|<\varepsilon$. Indeed, suppose for a contradiction that
there exist $\varepsilon>0$ and a sequence $(k_n) _{n\in\Bbb N}$, tending to infinity, such that for each $n$,
$$s(k_n)\in K_\epsilon=\{s\in K: |s(k_n)-s|\ge \varepsilon\mbox{ for each }s\in S\}.$$
A sequence $(s(k_n))_{n\in\Bbb N}$ of points of a compact set $K_\epsilon$ has a subsequence converging to a point $s\in K_\epsilon$. Since $\lim_{n \to \infty} N(s(k_n))=0$ and the function $N$ is continuous, we have $N(s)=0$. But then $s\in S$ and
so $|s(k_n)-s|<\varepsilon$ for some $n$. Then $s(k_n)\not\in K_\varepsilon$, a contradiction.
Update. Let $D(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_0$ and $N(s)=b_ms^n+b_{m-1}s^{n-1}+\dots+b_0$. Let $s(k)$ be a root of $f(k,s)$. If $N(s(k))=0$ then $D(s(k))=0$, so polynomials $N(s)$ and $D(s)$ have a common root, which is impossible. Therefore $N(s(k))\ne 0$ and $k=-D(s(k))/N(s(k))$.
Suppose that there exists a sequence $k_j$ tending to infinity such that $s(k_j)$ tends to infinity. Then  for each $j$,
$$k_j=-\frac{D(s(k_j))}{N(s(k_j))}=\frac{a_n s(k_j)^n+o(|s(k_j)|^n)} {b_m s(k_j)^m+o(|s(k_j)|^m)}=\frac{a_n}{b_m}s(k_j)^{n-m} +o(|s(k_j)^{n-m}|).$$
