Partial Derivatives Tend to Zero implies Limit to Infinity Exist? Let $g:\mathbb{R}^2\to\mathbb{R}$ be a function such that $\frac{\partial g}{\partial x}$ and $\frac{\partial g}{\partial y}$ exist and are continuous on $\mathbb{R}^2$.
Suppose that $$|\frac{\partial g}{\partial x}(x,y)|+|\frac{\partial g}{\partial y}(x,y)|\leq\frac{1}{x^2+y^2}$$ if $(x,y)\neq (0,0)$.
Prove that $\lim_{(x,y)\to\infty}g(x,y)$ exists.

I am not very sure how to go about proving this. I know that the first condition (partial derivatives exist and continuous) implies that $g$ is differentiable on $\mathbb{R}^2$.
Then, I am thinking of using Taylor's Theorem somehow. One problem I face is I don't know what is the candidate limit of $\lim_{(x,y)\to\infty}g(x,y)$ which makes it hard to prove it using the Epsilon-Delta method.
Another line of thinking could be the fundamental theorem of line integrals (as suggested by a commenter below). Again, I face problems coming up with the full proof.
Thanks for any help!
Updates (I am working on this question in the meantime):
If we are proving using the $\epsilon-\delta$ definition, we need to show that there exists $c\in\mathbb{R}$ such that for all $\epsilon>0$ there exists $R>0$ such that whenever $\sqrt{x^2+y^2}>R$ we have $|g(x,y)-c|<\epsilon$. Finding a candidate for $c$ seems very hard to me, perhaps the way is to prove by contradiction.
I am following AlexM.'s solution, which looks promising, except for some details in the inequality. I am thinking if that can be fixed using application of Mean Value Theorem for Definite Integrals.
 A: Line integrals seem like the way to go. Consider a circle of radius $R$ around the origin. Since
$$ \sqrt{\Bigl(\frac{\partial g}{\partial x}\Bigr)^2+\Bigl(\frac{\partial g}{\partial y}\Bigr)^2} \le \frac{1}{R^2} $$
on the circle, by integrating along the circle we find that the maximal possible difference between two points that both have norm $R$ is $\pi/R$ which goes to $0$ as $R\to\infty$.
Furthermore
$$ g(x,0) = g(1,0) + \int_1^x \frac{\partial g}{\partial x}(t,0)\, dt $$
tends to a limit for $x\to \infty$, because the improper integral
$$ \int_1^\infty \frac{\partial g}{\partial x}(t,0)\, dt $$
exists, by squeezing between $\int_1^\infty \frac{\pm 1}{t^2} \,dt $
These two facts combine to show that the value of $g$ on all rays away from the origin converge uniformly towards $\lim_{x\to\infty} g(x,0)$.
A: Showing that the limit is the same in all $\infty$ `directions':
Let $R \in \mathbb R_+$. We will take the limit of $R \rightarrow \infty$ and see that a circle of this radius around the origin contains an ever-shrinking range of values. This shows that the limit is the same in all directions.
For any curve $c(s)$ parametrised by its length, we know that
$$\begin{align}
\left| \frac{d g(c(s))}{ds} \right| &= \left| \frac{\partial g}{\partial x} \frac{d c_x(s)}{ds} + \frac{\partial g}{\partial y} \frac{d c_y(s)}{ds} \right|\\
& \leq \left| \frac{\partial g}{\partial x} \right| + \left| \frac{\partial g}{\partial y} \right| \tag{Parametrising by length}\\
& \leq \frac 1 {x^2 + y^2} \tag{Evaluated at $(x(s), y(s)) = (c_x(s), c_y(s))$}
\end{align}$$
By choosing a circle, we've made the job easy: $x^2 + y^2 = R^2$. Between any two points on a circle, the largest distance between them allows for the largest possible discrepancy in value, and this length is $\pi R$. An upper bound on the discrepancy in the value between points is 
$$|g(p_1) - g(p_2) | \leq \int_{p_1}^{p_2} \left| \frac{d g(c(s))}{ds} \right| ds \leq \int_{p_1}^{p_2} \frac 1 {R^2} ds \; \leq \frac{\pi R}{R^2} = \frac \pi R$$
Now of course, as $R\rightarrow \infty$, we can clearly see that this tends to zero. So the difference between the limits at infinity (if they exist) tends to zero, and so there is either one limit for any $\infty$ direction, or none at all.
Showing that there is a limit in an $\infty$ direction:
We can choose any path we like to find that there is a limit. For convenience, let's choose the path along the $x$ axis: $c(s) = (s, 0)$. Naively, we could use the same technique as before to say that the value as $s \rightarrow \infty$ is
$$\lim_{s \rightarrow \infty} g(0,s) = g(0,0) + \int_{0}^\infty \frac{\partial g}{\partial x}(0,x') d x'$$
which converges iff the integral converges. Does the integral converge? Well, we know something about its absolute convergence:
$$\int_{0}^\infty \left| \frac{\partial g}{\partial x}(0,x') \right| d x' \leq \int_{0}^\infty \left| \frac 1 {x'^2} \right| d x'$$
but this upper bounding integral doesn't converge! No matter. We can make it converge by starting a little to the right of the origin. Let's say $x=1$, though it really doesn't matter where. Then:
$$\lim_{s \rightarrow \infty} g(0,s) = g(0,1) + \int_{1}^\infty \frac{\partial g}{\partial x}(0,x') d x'$$
and we know that the integral converges absolutely because
$$\int_{1}^\infty \left| \frac{\partial g}{\partial x}(0,x') \right| d x' \leq \int_{1}^\infty \left| \frac 1 {x'^2} \right| d x' = 1$$
Absolute convergence implies convergence, and hence the limit exists.
