TITLE: Real Analysis sequence limit problem
QUESTION [5 upvotes]: Sorry for the unclear title, the problem is too specific so I couldn't think of anything else.
Here goes:

If 
  $$
  \lim_{n \to \infty} \frac{x_{n+1}}{x_n} = a > 0,
$$
  prove that
  $$
  \lim_{n \to \infty} \sqrt[n]{x_n} = a.
$$

Now, in my textbook there is a proof provided but I don't understand it.
It goes like this:
\begin{equation}
\tag{1}
    \sqrt[n]{x_n}
  = \sqrt[n]{
      \frac{x_n}{x_{n-1}}
      \times
      \frac{x_{n-1}}{x_{n-2}}
      \times
      \cdots
      \times
      \frac{x_2}{x_1}
      \times
      \frac{x_1}{1}
    }
\end{equation}
Then they take $\log$ of both sides:
\begin{equation}
\tag{2}
   \log \sqrt[n]{x_n}
 = \frac{1}{n}
   \left(
       \log \frac{x_n}{x_{n-1}}
     + \log \frac{x_{n-1}}{x_{n-2}}
     + \dotsb
     + \log \frac{x_2}{x_1}
     + \log \frac{x_1}{1}
   \right)
\end{equation}
These two steps are clear to me. What comes next is what I don't understand:
\begin{equation}
\tag{3}
    \lim_{n \to \infty} \log \sqrt[n]{x_n}
  = \log \lim_{n \to \infty} \frac{x_n}{x_{n-1}}
  = \log a
\end{equation}
The textbook provides no other explanation for this except for a little note saying “Cauchy's theorem”. The only Cauchy theorem previously mentioned in the textbook was the first one in this article.
Then the rest of the proof looks like this:
\begin{equation}
\tag{4}
    e^{\log \lim_{n \to \infty} \sqrt[n]{x_n}}
  = e^{\log a}
\end{equation}
\begin{equation}
\tag{5}
    \lim_{n \to \infty} \sqrt[n]{x_n}
  = a
\end{equation}
Where I also have no idea what's happening.
Any ideas?
Thanks.

REPLY [10 votes]: What's happenning is that if a sequence $a_n$ tends to a limit, then the sequence of averages $\frac{1}{n}\sum_{k=1}^na_k$ also tends to the same limit. This is Cauchy's first theorem of limits, according to the article you referred to.