how to translate local reciprocity with galois groups into local reciprocity with weil groups? I am trying to understand certain aspects of the Weil group $W_K$ for a $p$-adic field K, in particular how it does interplay with local class field theory. 
Let $L/K$ be a finite unramified extension of such fields. Then by local reciprocity we have an isomorphism
$$\operatorname{Gal}(L/K)\cong K^*/N_{K/L}(L^*),$$
where $N_{L/K}$ denotes the norm of the extension. Under this isomorphism the Frobenius element is mapped to the element $\pi_K(\mod N_{L/K}(L^*))$, where $\pi_K$ is a prime element of $K$. Note that the class $\pi_K(\mod N_{L/K}(L^*))$ does not depend on the choice of prime element $\pi_K$, because all units in $K$ are some norms coming from $L$ (not trivial to prove!). On the other hand, we also have the projection $K^*\longrightarrow K^*/N_{K/L}(L^*)$.
From here on, my goal is to arrive at the (Artin reciprocity) homomorphism
$$W_K\longrightarrow K^*$$
For this purpose I take the projective limit over all finite unramified extensions $L/K$. Thus, we have 
$$\operatorname{Gal}(K^{nr}/K)\cong\varprojlim_{L/K}K^*/N_{K/L}(L^*),$$
where $K^{nr}$ is the maximal non-ramified extension of $K$. Almost by definition of $W_K$, there is a (continuous) homomorphism $W_K\longrightarrow\operatorname{Gal}(K^{nr}/K)$. 
But how do we get a (continuous) canonical homomorphism 
$$\varprojlim_{L/K}K^*/N_{K/L}(L^*)\longrightarrow K^*$$?
My guess is, there is something easy that I don't know about projective limits...
EDIT: Per Matt's comment and answer, taking the projective limit over only finite unramified extensions does not suffice for constructing the desired homomorphism $W_K\longrightarrow K^*$.
 A: Local Artin reciprocity, for a finite abelian extension $L/K$ with $K$ a $p$-adic field,
is a specific isomorphism $K^{\times}/N_{L/K}(L^{\times}) \cong Gal(L/K)$.
You seem particular interested in the unramified case, but let me treat the
arbitrary case first.
Passing to the inverse limit over $L$, one gets an isomorphism
$$\varprojlim{} K^{\times}/N_{L/K}(L^{\times}) \cong G_K^{ab}.$$
As Mephisto notes in their answer, the norm groups range over all open
subgroups of $K^{\times}$, and so we may rewrite this as
an isomorphism
$$\widehat{K^{\times}} \cong G_K^{ab},$$
where $\widehat{K^{\times}}$ is the profinite completion of $K^{\times}$.
Recall that, if we choose a uniformizer $\pi$ for $K$,
then $K^{\times} \cong \mathcal O^{\times} \times \mathbb Z$,
where $\mathcal O$ denotes the ring of integers in $K$, and the isomorphism
is given by mapping an element $a \in K^{\times}$ to 
$\bigl(a/\pi^{v(a)}, v(a) \bigr),$ where $v: K^{\times} \to \mathbb Z$
is the valuation, normalized via $v(\pi) = 1$.
Thus $\widehat{K^{\times}} \cong \mathcal O^{\times} \times \hat{\mathbb Z}$.
(Recall that $\mathcal O^{\times}$ is its own profinite completion, 
but $\mathbb Z$ is not; we let $\hat{\mathbb Z}$ denote the profinite completion
of $\mathbb Z$.)
Now we can understand what happens if we restrict to unramified extensions.
The valuation $v: K^{\times} \to \mathbb Z$ induces a projection
$\widehat{K^{\times}} \to \hat{\mathbb Z}$, which is independent of the choice
of $\pi$.  Similarly, there is a surjection $G_K^{ab} \to Gal(K^{nr}/K)$.
The latter group is isomorphic to $\widehat{Z}$; it is naturally identified with
the absolute Galois group of the residue field, and is topologically generated
by Frobenius.
Under the reciprocity isomorphism
$\widehat{K^{\times}} \cong G_K^{ab}$, the projection to $\widehat{Z}$ on the
source and the projection to $Gal(K^{nr}/K)$ on the target are compatible
with the isomorphism $\widehat{Z} \cong Gal(K^{nr}/K)$ given by mapping $1$ to Frobenius.
Now we can bring in Weil groups.  The general definition of the Weil group is somewhat involved, involving the fundamental classes in $H^2(L/K, L^{\times})$,
but after you sort everything out, you find that the  Weil
group $W_K$ can be identified with a subgroup of $G_K$, namely the preimage
under the natural map $G_K \to Gal(K^{nr}/K) \cong \hat{\mathbb Z}$ of
the subgroup $\mathbb Z \subset \widehat{\mathbb Z}$.  From this, one sees
that $W_K^{ab}$ can be identified with the subgroup of $G_K^{ab}$ which again
is the preimage under the natural map $G_K \to Gal(K^{nr}/K) \cong \hat{\mathbb Z}$ of the subgroup $\mathbb Z\subset \widehat{\mathbb Z}$.
If we go back to the preceding discussion of our reciprocity map, we see that
the isomorphism $\widehat{K^{\times}} \cong G_K^{ab}$ restricts to an isomorphism $K^{\times} \cong W_K^{ab}$.  The inverse of this is the isomorphism you are looking for, I would guess.
If one restricts to the unramified case, then we have to pass to the quotient
$\mathbb Z$ of $K^{\times}$, and the quotient $\mathbb Z \subset \widehat{\mathbb Z} = Gal(K^{nr}/K)$ of $W_K^{ab}$, and then the isomorphism
just becomes $\mathbb Z = \mathbb Z$.
Additional remarks: There are various confusions in your question.  Here are some: you write $W_K$ but you mean $W_K^{ab}$.  (The Weil group itself is not
abelian, just as $G_K$ is not abelian.)  You say you want the general Artin
reciprocity isomorphism $W_K$ [sic] $\to K^{\times}$, but you then restrict
attention to unramified extensions.  As noted above, these will only see the quotient $\mathbb Z$ of $W_K^{ab}$ and the corresponding quotient $\mathbb Z$
of $K^{\times}$.   You then go on to ask for a continuous homomorphism
$\varprojlim{} K^{\times}/N_{L/K}(L^{\times}) \to K^{\times}.$  Leaving aside the issue that the left-hand side should involve all abelian $L$ over $K$,
not just the unramified ones (or else the left-hand side will be too small),
there is no such map, since the left-hand side is profinite and the right hand side has a discrete factor (as noted above).  The correct thing is to profinitely complete $K^{\times}$, and one then has the reciprocity isomorphism
discussed above.
Further remarks: Here are some additional remarks, prompted in part by the exchange of comments below.  
At a technical level, passing from $G_K$ to $W_K$ simply replaces the $\widehat{\mathbb Z}$ quotient of $G_K$ coming from the map $G_K \to Gal(K^{nr}/K) \cong \widehat{\mathbb Z}$ by $\mathbb Z$, so that instead of writing the reciprocity isomorphism as an isomorphism
$\widehat{K^{\times}} \cong G_K^{ab}$ we can rewrite it as an isomorphism
$K^{\times} \cong W_K^{ab}$, and so avoid passing from $K^{\times}$ to
$\widehat{K^{\times}}$.
As for why we do this: one reason is that $K^{\times}$ is what appears as a local factor in the ideles, not $\widehat{K^{\times}}$.  There are additional motivations.   One thing to remember is that the original definition of the
Weil group is not as the preimage of $\mathbb Z \subset \widehat{\mathbb Z}$
in $G_K$, but in terms of the fundamental classes of local class field theory;
so the Weil group arises naturally from this point of view.  For further motivations, this MO post might help.
A: You need to use that the norm groups are exactly the open subgroups of finite index in $K^*$ (existence theorem; the subgroups of finite index are automatically open in the characteristic zero case). Therefore the inverse limit is just the profinite completion of $K^*$, which is $K^*$ itself. [Correction: not quite; the quotient $\mathbb{Z}$ of $K^*$ has to be replaced by its profinite completion.] So the canonical homomorphism you want goes the other way.
I am assuming the limit is over all finite extensions $L$ of $K$. If you only take the unramified finite extensions then the inverse limit is equal to profinite completion of $\mathbb{Z}$
