Why does the amoeba shrink to its skeleton when we go to infinity? Let $f\in\mathbb{C}[X_1^{\pm1},\ldots,X_n^{\pm1}]$ a Laurent polynomial.
Let $\mathrm{Log}:(\mathbb{C}\setminus\{0\})^n\to\mathbb{R}^n$ defined by $\mathrm{Log}(z_1,\ldots,z_n)=(\log|z_1|,\ldots,\log|z_n|)$.
We call amoeba of $f$ the set $\mathcal{A}_f=\mathrm{Log}(f^{-1}(0))$, i.e. the image of the variety defined by $f$ under the logarithm.
It can be shown that there exists a convex subdivisions of $\mathbb{R}^n$ such that, each $n$-dimensional cell contains exactly one connected component of $\mathbb{R}^n\setminus\mathcal{A}_f$. Then the $(n-1)$-dimensional cells are contained in the amoeba and form the so called skeleton of the amoeba.
From the pictures (c.f. the links), it seems obvious that when we go to infinity the amoeba shrinks to its skeleton.
Because of some properties of the amoeba, what I want to show can be formulated in the following way:

If a convex polyhedron $P$ is contained in $\mathcal{A}_f$, then
either $P$ is bounded
or $P$ has empty interior.

And, if we write
$B(\boldsymbol{x},r)=\{\boldsymbol{y}\in\mathbb{R}^n \mid \|\boldsymbol{y}-\boldsymbol{x}\|<r\}$ the ball of center $\boldsymbol{x}$ and radius $r$; this is equivalent to:

For any $\epsilon>0$ there exists $D>>0$ such that
$$\|x\|>D\Rightarrow B(\boldsymbol{x},\epsilon)\not\subset\mathcal{A}_f.$$

In the article of Passare and Rullgard it is shown that the skeleton is a deformation retract of the amoeba, but this does not imply that the amoeba shrinks on its skeleton when we go to infinity.
An other nice reference on this subject.
Some more properties:

*

*The connected components of $\mathbb{R}^n\setminus \mathcal{A}_f$ are convex.

*The convex subdivision of $\mathbb{R}$ is dual to a convex subdivision of the newton polytope of $f$.

*Let $E$ be a connected component of $\mathbb{R}^n\setminus \mathcal{A}_f$ and $\sigma$ the $n$-cell that contains it. Then the recession cones of $E$ and $\sigma$ are the same.

 A: Here's a sketch of what's going on.
Near any$^{\star}$ large solution to $f(z_1, \ldots, z_n) = 0$, some of the monomial terms $\beta_{\vec{\alpha}} z_1 ^{\alpha_1}z_2^{\alpha_2}\ldots z_n^{\alpha_n} (\alpha_i \in \mathbb{Z},\; \beta_{\vec{\alpha}} \in \mathbb{C}\setminus\{0\})$ are large. But if the sum of all the monomials is zero at $\vec{z}$, then all the large terms need to cancel. In particular, we can't have one term of larger size than the sum of the moduli of all the others.
The Newton polytope of $f$ is the convex hull of the $\alpha$ featuring in $f$, and its vertices correspond to the monomials which can be of largest size at any $\vec{z} \in \mathbb{C}^n$. 
A face $C$ of the polytope determines a ray in $\mathbb{R^n}$ such that if $\mathrm{Log}(z)$ lies on the ray, for any $\vec{\alpha}$ on the face of the polytope, $\beta_\vec{\alpha}\vec{z}^{\vec{\alpha}}$ is of the same size. (If all the $\beta_\vec{\alpha}$ of these leading monomials for $\vec{z}$ are 1, then the ray passes through the origin, otherwise it is displaced). For shorthand write this $|\beta_\vec{\alpha}\vec{z}^{\vec{\alpha}}| =: |\vec{z}|^C$
Similarly an edge of the polytope determines a hyperplane in $\mathbb{R^n}$; the rays arise at the intersection of $n-1$ of these hyperplanes; and analogously for intermediate-dimensional edge cells of the polytope (and we'll call any of them $C$ too). The union of the hyperplanes is the skeleton of the amoeba of $\mathcal{A}_f$.
Now, the solutions to $f(\vec{z})=0$ do not in general have their $\mathrm{Log}$ lying exactly on this skeleton, for two reasons. First, because there is a contribution from the non-leading terms; and secondly, at least if there are more than two terms on the leading cell, their contributions may not be exactly equal, though we can be sure that no one of them overwhelms all the others.
First, let's consider the non-leading terms. All the monomials of $f$ that are not leading for $\vec{z}$ lie an orthogonal distance at least $q$ behind the leading edge (or face, or intermediate-dimensional cell, according to the choice of $\vec{z}$) for $\vec{z}$. 
In which case their maximum size is proportional to $(|\vec{z}|^C)^{1-q}$ for leading $\vec{z}$, and their total contribution is smaller than the maximum size of the $\beta$ times the number of monomials in $f$. 
As $|\vec{z}|$ increases, this contribution decreases as a proportion of the size of the leading terms $|\vec{z}|^C$. So the distance of the deviation from the skeleton of $\mathrm{Log}(\vec{z})$ needed to account for this contribution decreases: it is proportional to $(1-q) \log (|\vec{z}|^C)$.
Moreover, any deviation from the skeleton favours one side or other (relative to its defining edges) of the leading face $C$ and there is a similar constant power $q$ that determines how far we may go before those terms overwhelm the contribution of the other vertices of $C$; which again allows us only a distance $\propto (1-q) \log (|\vec{z}|^C)$ from the skeleton. 
[...]
The above is almost a proof of the desired result. The obvious gap is in the strict dichotomy between leading and non-leading terms, which results in ambiguity about what happens when we deviate from a point at which a face is leading towards a lower-dimensional leading cell, or vice versa. It can be tightened by considering in more detail the distance from the low-dimensional intersections of the skeleton (corresponding to higher-dimensional cells of the polytope) at which this ambiguity can occur. 
$\star$ Provided that the Newton polytope contains the origin; but we can multiply $f$ by a suitable (possibly rational) power of $X_1, \ldots, X_n$, to achieve this without loss of generality.
