Limit of polynomial division $\lim\limits_{n \to \infty} \frac{f(n)}{g(n)}$ I am asked to figure out when this limit exists for polynomials $f$ and $g$, if so what the limit is, and then to prove my findings. So far I have gathered that the limit exists if and only if $\deg f\leq \deg g$, in which case the limit is the division of the leading coefficients (if $\deg f = \deg g$) and otherwise $0$ (if $\deg g > \deg f$).
I am stuck on proving this rigorously with the definition of a limit. This is my first course in real analysis and I find that most of our problems have quite intuitive results but the proofs are super involved. Hints are appreciated.
 A: For rational functions, the limit as $x$ approaches infinity is based on the limit $$\lim\limits_{x\to\infty}\frac1{x}=0.$$
Suppose in the following function $f(x)$, the numerator is the sum of $ax^n$, the term with the highest exponent, and $\mathcal{A}(x)$, the rest of the terms of the polynomial, each with a term with an exponent lower than $ax^n$.  The same principle applies to the denominator.
Analyzing:
\begin{align}
f(x)&=\frac{ax^n+\mathcal{A}(x)}{bx^m+\mathcal{B}(x)}\\
\lim\limits_{x\to\infty}f(x)&=\lim\limits_{x\to\infty}\frac{ax^n+\mathcal{A}(x)}{bx^m+\mathcal{B}(x)}\\
&=\lim\limits_{x\to\infty}\frac{ax^n+\mathcal{A}(x)}{bx^m+\mathcal{B}(x)}\color{blue}{\cdot\frac{\frac{1}{x^m}}{\frac{1}{x^m}}}\\
&=\lim\limits_{x\to\infty}\frac{ax^{n-m}+\frac1{x^m}\mathcal{A}(x)}{b+\frac1{x^m}\mathcal{B}(x)}\\
&=\frac{\lim\limits_{x\to\infty}ax^{n-m}+\lim\limits_{x\to\infty}\frac1{x^m}\mathcal{A}(x)}{\lim\limits_{x\to\infty}b+\lim\limits_{x\to\infty}\frac1{x^m}\mathcal{B}(x)}\\
\end{align}
We can then split this into three scenarios:
For $n<m$:  $\lim\limits_{x\to\infty} f(x)=0$.
For $n=m$:  $\lim\limits_{x\to\infty}f(x)=\frac{a}{b}$.
For $n>m$:  $\lim\limits_{x\to\infty}f(x)$ diverges.
A: Hint: $\frac { \sum\limits_{k=0}^{p} a_kx^{k}} {\sum\limits_{k=0}^{m} b_kx^{k}}= x^{p-m} \frac { \sum\limits_{k=0}^{p} a_kx^{k-p}} {\sum\limits_{k=0}^{m} b_kx^{k-m}}$ and $\frac { \sum\limits_{k=0}^{p} a_kx^{k-p}} {\sum\limits_{k=0}^{m} b_kx^{k-m}} \to \frac {a_p} {b_m}$ as $x \to \infty$. 
So you only have to see what happens to $x^{p-m}$ as $x \to \infty$ and this is very easy. 
A: Let the expression be be $\sum_{i=0}^n a_ix^i\over\sum_{i=0}^mb_ix^i$.
We may rewrite this as $a_nx^n\bigl(1 + \sum_{i=1}^n \frac{a_i}{a_n}x^{-i}\bigr)\over b_mx^m\bigl(1 + \sum_{i=1}^m \frac{a_i}{a_m}x^{-i}\bigr)$
Now, consider the number $y =\frac1{\epsilon^\frac 1k}$ where $\epsilon$ and $k$ are any positive numbers. We will prove $\lim\limits_{x\to\infty} \frac 1{x^k} = 0$. Since $x \mapsto \infty$ $x > y$. So, $$x^k > \frac 1\epsilon$$
$$=> \frac 1{x^k} -0 < \epsilon$$
So, $x^{-k} - 0$ is less than any positive real number, which means $x^{-k} = 0$ as $x \mapsto \infty$. 
So as $x \mapsto \infty$ our expression simplifies to $\frac {a_n} {b_n} x^{n-m}$. With similar reasoning as above we reach the $conclusion$. 
