Suppose $F(x)$ is a continuously differentiable (that is, first derivative exist and are continuous) vector field in $\mathbb{R}^3$ that satisfies the bound $$|F(x)| \leq \frac{1}{1 + |x|^3}$$ Show that $\int\int\int_{\mathbb{R}^3} \text{div} F dx = 0$.
Attempted proof - Suppose $F(x)$ is a continuously differentiable vector field in $\mathbb{R}^3$ such that $$|F(x)| \leq \frac{1}{1 + |x|^3}$$ From defintion of vector field We have $F:D\subseteq \mathbb{R}^3\to \mathbb{R}^3$ where $D$ is an open subset of $\mathbb{R}^3$. For every $x\in \mathbb{R}^3$, we can write $$F(x) = F_1(x) i + F_2(x) j + F_3(x) k$$ where $ i,j,k$ are the unit vectors. Since $F(x)$ is continuously-differentiable then so are the component functions.
I am a bit lost on trying to go from here. I know that $$\text{div} F = \nabla F = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\cdot \left(F_1,F_2,F_3\right) = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}$$ I think I need to incorporate the bound that $|F(x)|$ in order to show the desired result. Perhaps this is best done using an epsilon-delta sort of proof but I am not sure. Any suggestions are greatly appreciated.
Attempted proof 2 - Let $\{B(0,n)\}_{n=1}^{\infty}$ be a sequence of balls. Since $F(x)$ is continuously differentiable vector field, its components are also continuously differentiable. We know that the surface area of $\partial B(0,n)$ is $4\pi n^2$ and $$|F|\leq \frac{1}{1 + n^3}$$ Thus from the divergent theorem we have $$\int\int\int_{\mathbb{R}^3}\text{div}F dx = \int\int_{B(0,n)}F \times 0 = 0$$
Attempted proof 3 - Let $\{B(0,n)\}_{n=1}^{\infty}$ be a sequence of balls. Since $F(x)$ is a continuously differentiable vector field, its components are also continuously differentiable. We know that the surface area of $\partial B(0,n)$ is $4\pi n^2$ and $$|F|\leq \frac{1}{1 + n^3}$$ Let $S$ be the boundary surface of $B(0,n)$ with positive orientation then from the divergence theorem $$\iint_{S}F dS = \iiint_{B(0,n)}F dB(0,n) $$ Thus if we let $n\to \infty$ we see that $$\iiint_{B(0,n)}F dB(0,n) = 0$$
Attempted proof 4 - Let $\{B(0,n)\}_{n=1}^{\infty}$ be a sequence of balls. Since $F(x)$ is continuously differentiable vector field, so are its components. Let $S$ be the boundary surface of $B(0,n)$ with positive orientation. We know that the surface area of $\partial B(0,n)$ is $4\pi n^2$ and that $$|F|\leq \frac{1}{1 + n^3}$$ Thus if we apply the divergence theorem on the sequence of balls then we have $$\Big|\int_{\mathbb{R}^3}\text{div}F\,dx\Big| = \lim_n\Big|\int_{B(0,n)}\text{div}F\,dx\Big| = \lim_n\Big|\int_{\partial B(0,n)}F \cdot \nu\,dS\Big| \le \lim_n\frac{4\pi n^2}{1 + n^3} = 0.$$ Thus the result follows.
Apply the divergence theorem on a sequence of balls $B(0,n)$ where $n\to\infty$ and use the bound on $F$ to bound the boundary integral $\int_{\partial B(0,n)}F\cdot \nu\,dS$