Understanding the definition of the order of an entire function in Ahlfors's Complex Analysis Let $f: \mathbb C \to \mathbb C$ be an entire function. The order of $f$ is defined by $$\lambda=\limsup_{r \to \infty} \frac{\log \log M(r)}{\log r}, $$ where $$M(r)=\max_{|z|=r} |f(z)| .$$
Ahlfors in his Complex Analysis claims that

"According to this definition $\lambda$ is the smallest number such that
      $$M(r)\leq e^{r^{\lambda+\varepsilon}} $$ for any given $\varepsilon > 0$ as soon as $r$ is sufficiently large."

Why is this true?
My attempt:
We know that $$\lambda=\lim_{\rho \to \infty} \sup_{r \geq \rho} \frac{\log \log M(r)}{\log r}. $$
From the definition of the limit we have that for any $\varepsilon>0$, there exists some $\rho_0>0$, such that $$\left\lvert \sup_{r \geq \rho} \frac{\log \log M(r)}{\log r}-\lambda \right\rvert \leq \varepsilon ,$$ for every $\rho \geq \rho_0$. In other words $$\frac{\log \log M(r)}{\log r} \leq \lambda+\varepsilon $$ for every $r \geq \rho_0$. From here it is easy to see that $$M(r)\leq e^{r^{\lambda+\varepsilon}}, $$ for all $r \geq \rho_0$. I cannot see why $\lambda$ is the smallest number with this property.
Thanks in advance.
 A: Note that
$$F(\rho) = \sup_{r \geqslant \rho} \frac{\log\log M(r)}{\log r}$$
is a non-increasing function of $\rho$, hence $\lim\limits_{\rho\to\infty} F(\rho) = \inf\limits_{\rho > R} F(\rho).$
By the definition of the limes superior, for every $\mu < \lambda$, with $\varepsilon = \frac{\lambda-\mu}{3}$ there are arbitrarily large radii $r$ with
$$\frac{\log \log M(r)}{\log r} > \lambda - \varepsilon = \mu + 2\varepsilon,$$
and that inequality is equivalent to
$$M(r) > e^{r^{\mu+2\varepsilon}},$$
so for every $\mu < \lambda$, there is an $\varepsilon > 0$, such that there is no $\rho_0$ with
$$M(r) \leqslant e^{r^{\mu+\varepsilon}}$$
for all $r \geqslant \rho_0$, indeed
$$M(r)e^{-r^{\mu+\varepsilon}}$$ is unbounded for all small enough $\varepsilon > 0$ then.
A: I once stumbled upon this definition also. I would like to write up a little bit more details than Daniel's nice answer here. 
For convenience, define (${\small \text{we will discuss this definition later}}$) for $r>0$ that
$$
G(r):=\frac{\log\log M(r)}{\log r}.
$$
We prove the following proposition:

With the definition
  $
\lambda:=\limsup_{r\to\infty}G(r),
$
  the following hold.
(1) For every $\epsilon>0$, there exists $R>0$ such that for every $r\geqslant R$, 
      $$
  M(r)\leqslant \exp(r^{\lambda+\epsilon})\tag{*}
$$
  (2) If $\mu>0$ is such that for every $\epsilon>0$, there exists $R>0$ such that for every $r\geqslant R$, 
      $$
  M(r)\leqslant \exp(r^{\mu+\epsilon}),
$$
      then $\lambda\leqslant\mu$. (Note that this is what "smallest" means in Ahlfors.)

Proof. By definition of the $\limsup$ we have $$\lambda=\inf_{R>0}\sup_{r\geqslant R}G(r),$$ which implies (by the definition of $\inf$) that for every $\epsilon>0$, there exists $R>0$ such that 


*

*(i) $\sup_{r\geqslant R}G(r)<\lambda+\epsilon$;

*(ii) $\sup_{r\geqslant R}G(r)\geqslant\lambda$.


We will now prove that (i) implies (1) and (ii) implies (2). 
Note that (i) in particular implies that 
$$
G(r)<\lambda+\epsilon,\quad r\geqslant R.
$$
Taking the exponential function twice on both sides of
$$
\log(r)G(r)<\log(r)(\lambda+\epsilon)
$$
we get (1). 
On the other hand, the assumption of (2) implies that for every $\epsilon>0$,
$
\sup_{r>R}G(r)\leqslant \mu+\epsilon,
$
and thus
$$
\sup_{r>R}G(r)\leqslant \mu,
$$
which together with (ii) yields
$$
\lambda\leqslant\mu.
$$

Note: $G(r)$ is well-defined only for $M(r)>1$. But this is true for large enough $r$ since one could assume that the entire function is not constant. 
