Cake cutting problem This problem is the "piece of cake" problem from Elements of Information Theory for which there are many solutions available. The solution below is from this course.

A cake is sliced roughly in half, the largest piece being chosen each
  time, the other pieces discarded. We will assume that a random cut
  creates pieces of proportions:
$$P = \begin{cases} 
       (\frac{2}{3}, \frac{1}{3}) \;\; \text{with probability} \;\; \frac{3}{4}\\
       (\frac{2}{5}, \frac{3}{5}) \;\; \text{with probability} \;\; \frac{1}{4}    \end{cases}$$ Thus, for example, the first cut (and
  choice of largest piece) may result in a piece of size $\frac{3}{5}$.
  Cutting and choosing from this piece might reduce it to size
  $(\frac{3}{5})(\frac{2}{3})$ at time 2, and so on. How large, to first
  order in the exponent, is the piece of cake after $n$ cuts?

Solution: Let $C_i$ be the fraction of the piece of cake that is cut at the $i$th cut, and let $T_n$ be the fraction of cake left after
 $n$ cuts. Then we have $T_n = \Pi^n C_i$. Hence: $$\lim_{n \rightarrow
 \infty}\frac{1}{n} \log T_n = \lim_{n \rightarrow
 \infty}\frac{1}{n}\sum^n_i \log C_i \\
      = \mathbb{E}\log C_i \\
= \frac{3}{4}\log\frac{2}{3} + \frac{1}{4}\log\frac{3}{5} $$
Therefore after $n$ cuts, the size of the piece of cake is: $T_n =
 2^{n(\frac{3}{4}\log\frac{2}{3} + \frac{1}{4}\log\frac{3}{5})}$
My questions are: 


*

*On the first line, where did the $\lim_{n} \frac{1}{n}$ come from? 

*Why are we taking a limit at the beginning to arrive at that final log expression, and then plugging it right back in at the end to get our expression for $T_n$ which has an $n$? 


Note : all logs are to base 2
 A: You first have $$\log T_n = \sum_{i=1}^n \log C_i$$ (by definition of $T_n$ and the $C_i$'s, taking the logarithm on both sides of $T_n = \prod_{i=1}^n C_i$). From this, you "artificially" divide by $\frac{1}{n}$ on both sides, to make it look like an average over the $C_i$'s:
$$
\frac{1}{n}\log T_n = \frac{1}{n}\sum_{i=1}^n \log C_i
$$
Since all choices $C_i$'s are independent and identically distributed ($C_i = \frac{2}{3}$ w.p. 3/4, and $\frac{3}{5}$ w.p. 1/4), you can see them as i.i.d. realizations of a random variable $C$ with this distribution. By the law of large numbers, by taking the limit of the average of the first $n$ of the $\log C_i$'s,the RHS converges to the expectation of a random variable $\log C$, for $C$ with this distribution.
$$
\lim_{n\to\infty}\frac{1}{n}\log T_n = \lim_{n\to\infty} \frac{1}{n}\sum_{i=1}^n \log C_i = \mathbb{E}[\log C].
$$
But now, $\mathbb{E}[\log C]$ is easy to compute explicitly, since you fully now the (simple) distribution of $C$:
$$
\log C = \begin{cases}
\log \frac{2}{3} & \text{ w.p. } \frac{3}{4}\\
\log \frac{3}{5} & \text{ w.p. } \frac{1}{4}
\end{cases}
$$
and therefore
$$
\lim_{n\to\infty}\frac{1}{n}\log T_n = \mathbb{E}[\log C] = \frac{3}{4}\log \frac{2}{3} + \frac{1}{4}\log \frac{3}{5},
$$
as stated in the solution you linked.
Now, this gives us the limit $\ell=\frac{3}{4}\log \frac{2}{3} + \frac{1}{4}\log \frac{3}{5} \neq 0$ of the quantity
$$
\frac{1}{n} \log T_n.
$$
This is not what we want. However, this implies that asymptotically
$$
\log T_n = n\ell + o(n)
$$
when $n\to\infty$ (as $\frac{1}{n}\log T_n = \ell + o(1)$). Exponentiating, we get:
$$
T_n = 2^{n\ell + o(n)},
$$
which is the result we wanted ("first order in the exponent" approximation).
