The negation of a limit that tends to infinity. In my independent study course, I'm reading on entire transcendental functions and I'm faced with a minor issue.
This is a theorem in the book I have to read, I understood all of the proof but the start.
Theorem 9.1. If $f(z)$ is an entire transcendental function, with maximum modulus function $M(r)$, then
$$\lim_{r \to \infty} \frac{\ln M(r)}{\ln r}=\infty$$
The proof starts with supposition of the contrary, i.e., suppose that
$$\liminf_{r \to \infty} \frac{\ln M(r)}{\ln r}=\mu<\infty$$
And this is the part where I am confused with - why is a finite limit inferior the negation of the infinite limit? Can someone kindly explain this to me?
 A: For a function $f\colon \mathbb R\to\mathbb R$ (or similarly even just defined on a suitable smaller domain such as $(0,\infty)$) $\lim_{r\to \infty }f(r)=\infty$ means that for all $a\in\mathbb R$ there exists $R\in\mathbb R$ such that $r>R$ implies $f(r)>a$.
The negation says that there exists some $a$ such that for each $R$ there exists $r>R$ with $f(r)\le a$, consquently $\liminf_{r\to\infty} f(r)\le a$ for this $a$. Thus $$\neg\left(\lim_{r\to\infty}f(r)=\infty\right)\implies \liminf_{r\to\infty}f(r)<\infty$$
Of course we also have $\lim_{r\to\infty} f(r)=\infty\implies \liminf_{r\to\infty}f(r)=\infty$ so that indeed
$$\neg\left(\lim_{r\to\infty}f(r)=\infty\right)\iff \liminf_{r\to\infty}f(r)<\infty$$

To put it differently: Since the limit need not exist, the negation should deal both with nonexistent and with finite limits. However, for the special case of $\infty$, we have that $\lim=\infty$ implies $\liminf=\limsup=\infty$ whereas $\liminf=\infty$ leaves no room above it, so $\liminf=\limsup$ and hence $\lim$ exists (and equals $\infty$).
