Suppose $n=m$, as $x$ becomes larger and larger $f$ becomes closer to $1$ (Though I don't know how to show this rigorously). $$\forall \epsilon>0\exists N>0s.t(\forall x\in\mathbb{R})[x>N\longrightarrow |(a_nx^n+...+a_0)/(b_mx^m+...b_0)-l|<\epsilon]$$By letting $l=1$ ,for any $\epsilon>0$ one can choose a large enough $N$ such that $|(a_nx^n+...+a_0)/(b_mx^m+...b_0)-1|<\epsilon$ holds true.
Now suppose $m>n$ then as $x$ becomes larger and larger $f$ becomes closer to $0$ (Again don't know how to show this rigorously). Once more we have $$\forall \epsilon>0\exists N>0s.t(\forall x\in\mathbb{R})[x>N\longrightarrow |(a_nx^n+...+a_0)/(b_mx^m+...b_0)-l|<\epsilon]$$But this time by letting $l=0$ ,for any $\epsilon>0$ one can choose a large enough $N$ such that $|(a_nx^n+...+a_0)/(b_mx^m+...b_0)|<\epsilon$ holds true.
The converse argument shows why if $n>m$, $\lim_{x \to \infty } f(x)$ must not exist. Q.E.D?
Where I need help:I have two principal concerns, firstly I don't know how to structure my proof in order to satisfy the condition "if and only if". Secondly I feel as if my proof is missing rigor because I cannot prove that if $m=n$, $f$ tends to $1$ as $x$ becomes larger and if $m>n$$f$ tends to $0$. Any help will be appreciated, thank you.