Let $f(r)$ be the number of integral points inside circle of radius $r$ and center at origin,then $\lim_{r\to \infty}\frac{f(r)}{\pi r^2}$ Let $f(r)$ be the number of integral points inside circle of radius $r$ and center at origin,then $\lim_{r\to \infty}\frac{f(r)}{\pi r^2}$

I know the formula for number of lattice points inside the boundary of a circle of radius $r$ with center at the origin is given by $f(r)=1+4\lfloor r\rfloor+4\sum_{i=1}^{\lfloor r\rfloor}\lfloor \sqrt{r^2-i^2}\rfloor$
But i am not able to find $\lim_{r\to \infty}\frac{f(r)}{\pi r^2}$.
 A: Let $r$ be large. If $P$ is any lattice point inside the circle, colour blue the $1\times 1$ square which has $P$ as its centre and whose sides are parallel to the coordinate axes. Then the area $A$ coloured blue is the number of lattice points. 
Note that every point in the circle with centre the origin and radius $r-10$ is coloured blue, and every point that is coloured blue is in a circle of radius $r+10$ with centre the origin. (We can do better than $10$.) It follows that
$$\pi(r-10)^2\lt A\lt \pi(r+10)^2.$$
Thus 
$$\frac{\pi(r-10)^2}{r^2}\lt \frac{A}{r^2}\lt \frac{\pi(r+10)^2}{r^2}.$$
Now let $r\to\infty$.
A: Let $g(r)$ be the maximum number of squares of side 1 having integral vertices contained inside the circle of radius $r$.
Let $G(r)$ be the minimum number of squares of side 1 having integral vertices that contain the circle of radius $r$.
Consider a bijective mapping of every such square onto the lower left vertex.
Then simply counting all of them will give you the inequality,
$$g(r)<f(r)<G(r)$$
When we use second order integrals to find out the area of a figure we essentially divide it into infinitesimally small squares.This is exactly what we have done here. Hence, $\lim_{r\rightarrow \infty} \dfrac{g(r)}{\pi r^2}=\lim_{r\rightarrow \infty} \dfrac{G(r)}{\pi r^2}=1$
By sandwich theorem $\lim_{r\rightarrow \infty} \dfrac{f(r)}{\pi r^2}=1$
A: Using Euler-Mclaurin summation formula we can get that: $$f(r)=\pi r^2+O(r).$$ Hence $$\lim \limits_{r\to \infty}\dfrac{f(r)}{\pi r^2}=1.$$
