Confusion about limit of log I have a very elementar question but I do not see where my mistake is. 
Suppose we have a sequence $(x_n)$ with
$\lim_{n\to\infty}x_n=1$. Moreover, suppose that the sequence $({x_n}^c)$ for some constant $c>1$ has limit $\lim_{n\to\infty}{x_{n}}^c=c$.
Then
$$
\lim_{n\to\infty}\log({x_n}^c)=\log(c).
$$
But since $\log({x_n}^c)=c\log(x_n)$, I also have
$$
\lim_{n\to\infty}\log({x_n}^c)=c\lim_{n\to\infty}\log(x_n)=0.
$$
Where is my mistake? Maybe in the assumptions of the sequences.
 A: If $x_n\rightarrow 1$, the only way to get $x_n^c\rightarrow c$ is if $c=1$. There is no other exponent $c$ which makes this work. 
After all, if $x_n\rightarrow 1$, then $x_n^c \rightarrow 1^c = 1$ as well, by continuity (see below). But then if $1=\lim_{n\rightarrow\infty}x_n^c = c$, then necessarily $c=1$.
So you can't simultaneously have $c>1$ and $x_n^c \rightarrow c$ because this is a contradiction; anything derived from it may also be a contradiction. In particular, by continuity, $\lim_{n\rightarrow \infty}\log(x_n^c) = c \log(1) = 0$ always. But if we assume $\lim_{n\rightarrow\infty} x_n^c = c$ and $c>1$, we find that instead $\lim_{n\rightarrow \infty}\log(x_n^c) = \log(c) > \log(1) = 0$— a contradiction.
We assumed something that cannot happen, and got a contradiction as a result. 



*

*First, note that if $h$ is any continuous function, then $$\lim_{n\rightarrow \infty }h\left(x_n\right) = h\left(\lim_{n\rightarrow \infty }x_n\right)$$

*Next, let $f$ be the function $f(x) = x^c$, let $g(x)=\log(x)$, and suppose $x_n$ is a sequence such that $x_n \rightarrow 1$.

*Because $f$ and $g$ are continuous, we know that :
$$\lim_{n\rightarrow \infty} \log(x^c) = \lim_{n\rightarrow \infty} g(f(x_n)) = g\left(f\left(\lim_{n\rightarrow \infty} x_n\right)\right) = g(f(1)) = \log(1^c) = \log(1) = 0$$

*Because of the property of logs, we know that:
$$\lim_{n\rightarrow \infty} \log(x^c) = \lim_{n\rightarrow \infty} c\log(x_n) = c\lim_{n\rightarrow \infty}\log(x_n) = c \lim_{n\rightarrow \infty}g(x_n) = c \cdot g\left(\lim_{n\rightarrow \infty}x_n\right) = c\cdot g(1) = c\cdot \log(1) = c\cdot 0 = 0$$



Taking a step back, we have that $x_n\rightarrow 1$, and because of continuity, $x_n^c \rightarrow 1$ as well. Hence $\log(x_n)$ and $\log(x_n^c)$ both go to 0 as $n\rightarrow \infty$.
A: If $\lim\limits_{n\,\to\,\infty} x_n =1$ then $\lim\limits_{n\,\to\,\infty} x_n^c = 1.$
\begin{align}
\lim_{n\,\to\,\infty} (x_n^c) & = \left( \lim_{n\,\to\,\infty} x_n \right)^c & & \text{because } x \mapsto x^c \text{ is a continuous function} \\[10pt]
& = 1^c = 1.
\end{align}
