Exchanging limits with norms and linear functionals In a normed vector space $X$, when can we say:
$\lim\|x_n\|=\|\lim x_n\|$
and further, if $f\in X^{*}$, when can we say:
$\lim fx_n=f(\lim x_n)$?
 A: You can say that if $\lim x_n= x$, which is a particular case of the following

Theorem: Let $X$ and $Y$ be normed spaces. If $\lim x_n= x$ in $X$ and $T:X\to Y$ is continuous, then
  $$\lim T(x_n)=T(\lim x_n).$$

Proof: Let $\varepsilon>0$. As $T$ is continuous, there exists $\delta>0$ such that
$$\|x-y\|<\delta\quad\Longrightarrow\quad\|T(x)-T(y)\|<\varepsilon.$$
As $\lim x_n=x$, there exists $n_0\in\mathbb{N}$ such that
$$n>n_0\quad\Longrightarrow\quad\|x_n-x\|<\delta.$$
It follows that
$$n>n_0\quad\Longrightarrow\quad\|T(x_n)-T(x)\|<\varepsilon$$
and thus
$$\lim T(x_n)=T(x)=T(\lim x_n).$$

Corollary 1: Let $X$ and $Y$ be normed spaces. If $\lim x_n= x$, then
  $$\lim \|x_n\|=\|\lim x_n\|.$$

Proof: The mapping
$$\begin{align} T:X&\longrightarrow \mathbb{R}\\
x&\longmapsto \|x\|
\end{align}$$
is continuous.

Corollary 2: Let $X$ and $Y$ be normed spaces. If $\lim x_n= x$ and $f\in X^*$, then
  $$\lim f(x_n)= f(\lim x_n).$$

Proof: By definition of $X^*$, $f$ is continuous.
Remark: The theorem above is also true for more general spaces (e.g. metric spaces).
A: Assuming that the definition of $\lim_n x_n=x$ is: $\lim_{n}\|x_n-x\|=0$. Yes, this follows from the fact that limits can be exchanged with continuous functions. Specifically, $f(x)=\|x\|$ is continuous in $x$:
$$|f(x)-f(y)|=|\|x\|-\|y\||\leq \|x-y\|,$$
by the reverse triangle inequality. You can show this follows in a similar way for $f\in X^*$ in general. 
