Integral points inside a circle of radius r and centre at origin Let $f(r)$ be the number of integral points inside a circle of radius $r$ and center at the origin (an integral point is a point both of whose coordinates are integers). Then $\lim_{r\to\infty}\frac{f(r)}{r^2}$ is equal to
(A) $1$
(B) $\pi$
(C) $2\pi$
(D) $\pi/2$
I could not find a general formula to estimate number of points. Any ideas ? 
 A: Any lattice point $(j,k)\in{\mathbb Z}^2$ is the center of a square
$$Q_{jk}:=\left\{(x,y)\>\biggm|\>j-{1\over2}\leq x\leq j+{1\over2}, \ k-{1\over2}\leq y\leq k+{1\over2}\right\}$$
of area $1$, and different such squares have at most an edge in common. It follows that the area of a union of different such squares is equal to the number of the occurring squares.
Let an $r>1$ be given, and let $$L_r:=\bigl\{(j,k)\>\bigm|\> (j,k)\in B_r\bigr\}$$
be the set of lattice points contained in the closed disk of radius $r$ centered at $(0,0)$. Then the number $\#L_r$ of elements in $L_r$ satisfies 
$$\#L_r={\rm area}\left(\bigcup_{(j,k)\in L_r}Q_{jk}\right)\leq\pi\left(r+{1\over\sqrt{2}}\right)^2\ ,$$
since all $Q_{jk}$ involved here fit into a disk of radius $r+{1\over\sqrt{2}}$. On the other hand,
$$\#L_r={\rm area}\left(\bigcup_{(j,k)\in L_r}Q_{jk}\right)\geq\pi\left(r-{1\over\sqrt{2}}\right)^2\ ,$$
since the same $Q_{jk}$ cover at least the disk of radius $r-{1\over\sqrt{2}}$. The squeeze theorem then shows that in fact
$$\lim_{r\to\infty}{\#L_r\over r^2}=\pi \ ,$$
hence answer (B) is correct.
By the way: Finding the optimal error estimate for the approximation $\#L_r\doteq \pi r^2$ is one of the most famous (unsolved) problems in geometric number theory.
