The relationship between Hilbert polynomial and the degree of the variety In the note I'm reading there is the following theorem:

Let $X\subset \mathbb{P}^n$ be a projective variety, and $p_X$ is its Hilbert polynomial. If the dimension of $X$ is $n$, then the leading term of $p_X(m)$ is
  $$
\frac{\operatorname{deg}X}{n!}m^n
$$

However, I tried to test it using the variety $X=\mathbb{V}(x_0)\cup\mathbb{V}(x_1,x_2)\subset\mathbb{P^2}$, and found some problems.
We have $\mathbb{I}(X)=\langle x_0\rangle\cap\langle x_1,x_2\rangle=\langle x_0x_1,x_0x_2\rangle$, thus
$$
S(X)=k[x_0,x_1,x_2] / \mathbb{I}(X)\cong k[x_0]\oplus k[x_1,x_2]
$$
Thus the Hilbert function is 
$$
h_X(m)=\operatorname{dim}_k S(X)_{m}=1+(m+1)=m+2
$$
So the Hilbert polynomial is $p_X(m)=m+2$, which indicates that the degree of $X$ is $1$, because the dimension of $X$, which is the maximal dimension of its irreducible component, is $1$.
However, the geometric definition of degree is the maximal number of intersections between $X$ and a hyperplane in general position with dimension equal to the codimension of $X$, in this case $1$. Since $X$ is just the union of $x_0$-axis and the $(x_1,x_2)$-plane, one can easily see the degree of it should be 2.
Could any tell me what goes wrong here?

Updated: following is my geometric intuition: The line $L\in\mathbb{P}^2$ intersects $X$ at two points: $L_1$ and $L_2$(which is the $x_0$-axis).

 A: The calculation is: If $I_1 = <X_0>$ and $I_2 = <X_1, X_2>$ then we have $Z = Var(I)$ with $I = I_1 \cap I_2 = <X_0*X_1, X_0 * X_2>$ and $I_1 + I_2 = <X_0, X_1, X_2>$. To compute the homogenoues coordinate ring $S/I, S := k[X_0, X_1, X_2],$ we use the exact sequence
$$0 \longrightarrow S/I \longrightarrow S/I_1 \coprod S/I_2 \longrightarrow S/(I_1 + I_2) \longrightarrow 0$$
We have $S/(I_1 + I_2) = k$, hence $Hilb_{S/(I_1 + I_2)} = 0.$ Moreover $Hilb_{S/I_1}(t) = Hilb_{k[X_1,X_2]}(t) = t+1$. And $Hilb_{S/I_2}(t)=Hilb_{k[X_0]}(t) = 1$
Because the Hilbert polynomial is additive on exact sequences we get
$$Hilb_{S/I} = Hilb_{S/I_1} + Hilb_ {S/I_2} – Hilb_k$$
i.e.
$Hilb_{S/I}(t) = t+2-0 = t+2$ and Z has $degree = 1.$
This result does not contradict intuition: The general line in $\mathbb P^2$ does not pass through the distinguished point of $Z$. All lines in $\mathbb P^2$ form the dual $\mathbb P^2$, those passing through the distinguished point form a 1-dimensional subset, isomorphic to $\mathbb P^1.$   
