Note that

$$F(\rho) = \sup_{r \geqslant \rho} \frac{\log\log M(r)}{\log r}$$

is a non-increasing function of $\rho$, hence $\lim\limits_{\rho\to\infty} F(\rho) = \inf\limits_{\rho &gt; R} F(\rho).$

By the definition of the limes superior, for every $\mu &lt; \lambda$, with $\varepsilon = \frac{\lambda-\mu}{3}$ there are arbitrarily large radii $r$ with

$$\frac{\log \log M(r)}{\log r} &gt; \lambda - \varepsilon = \mu + 2\varepsilon,$$

and that inequality is equivalent to

$$M(r) &gt; e^{r^{\mu+2\varepsilon}},$$

so for every $\mu &lt; \lambda$, there is an $\varepsilon &gt; 0$, 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 $\varepsilon &gt; 0$ then.

Understanding the definition of the order of an entire function
I'm new to working with RAW images, and I'm capturing simultaneous RAW+JPGs with my new Lumix LX5, and using Bibble to view/process the results.

I'm very surprised that the RAW images taken at 24mm wide 16x9 seem to capture a different (and larger) sensor area compared to the JPGs. The RAW images seem to contain the equivalent of about 100 extra pixels on left and right sides, and a smaller number top and bottom. I say "equivalent", because the actual pixel counts of RAW and JPG are only slightly different, which implies some resizing must be going on...?

JPG: 3968 x 2232
RAW: 3976 x 2238

I guess this small difference is because JPG images must be 16x16 multiples>

The raw image displays noticeable vignetting in the extra pixels, and there's a fair bit of chromatic aberration. I can crop off the 'extra' pixels, but then my RAW image has fewer pixels in it than the JPG, which doesn't feel right. 

I'll try and add samples shortly.
