limit of sequence of circles in the plane Consider the sequence $ \displaystyle{ A_n = \{ (x,y) \in \mathbb{R} ^2 : (x-n)^2 + y^2 \leq n^2 \} ; \quad n \in \mathbb{N} }$ of circles.
Find the limit $ \displaystyle { \lim_{n \to \infty} A_n }$.
The only thing I can see that the sequence $ \displaystyle{ (A_n)}$ is increasing with drawing the circles, but I can't prove it.
Any ideas?
Thank's in advance!
edit: I made a typo. It is $n^2$ the right side. Sorry for the confusion.
 A: If you replace $n$ with $n^{2}$, this is a sequence of circles that at each stage is increased in radius by $1$ and then shifted to the right by $1$. I claim that $\displaystyle\lim_{n \to \infty} A_{n}$ is actually the right-half plane with the origin, i.e. 
$$ \{(x,y) \in \mathbb{R}^{2}: x > 0 \} \cup \{0,0\}$$
It is easy to see that 
$$ \{(x,y) \in \mathbb{R}^{2}: x <0 \} \cup \{(0,y): y \neq 0\} $$
is not in $\displaystyle\lim_{n \to \infty} A_{n}$. Take $(x,y)$ with positive $x$ coordinate. We wish to find an $N$ large enough so that $(x,y) \in A_{n}$ for $n \geq N$. Without loss of generality, we assume $y$ is positive. 
Since the ball is given by
$$(x-n)^2 + y^{2} \leq n^2$$
we can solve for $y$, to get
$$y \leq \sqrt{n^{2} - (x-n)^{2}}$$
since $x > 0$, 
$$n^{2} - (x-n)^{2} > 0$$
Moreover, its derivative is $2(n-x)$, so that if $n > x$, 
$$\sqrt{n^{2} - (x-n)^{2}}$$
is increasing. Thus, taking $N$ larger than $x$ such that
$$y \leq \sqrt{N^{2} - (x-N)^{2}}$$
holds, we see that $(x,y) \in A_{n}$ for $n \geq N$. 
A: $A_n$ is a circle centred at $(n,0)$ with radius $\sqrt{n}$.  Since $n -\sqrt{n} \to +\infty$, there isn't a finite limit; the circles march off toward infinity along the positive $x$ axis.  
Perhaps you meant the right side to be $n^2$ instead of $n$.  That would make things more interesting, as all points on the $y$ axis would be limit points of this sequence of circles.
A: These circles are centered at $(n,0)$, so the center is not converging. It's moving toward $(\infty,0)$. The radius of the $n$th circle is $n$, so the radius is diverging as well, but left-most point of the circle is always $(0,0)$.
Take any $(x,y)$ in the right half-plane. Consider the line segment from the origin to this $(x,y)$. Its perpendicular bisector will intersect the $x$-axis at some point, say $(X,0)$. Now for all $n>X$, $(x,y)$ is within $A_n$. So every point in the right half-plane is in the limit of $\left\{A_n\right\}$. 
Considering a sequence of $(x_n,y_n)$ in the right half-plane that converges to $(0,y)$ on the $y$-axis, every point on the $y$-axis is also in the limit of $\left\{A_n\right\}$.
So $\lim A_n=\left\{(x,y)\mid x\geq0\right\}$.
A: The circles are centered at $(n,0)$ with a radius of $n$. Considering this you can decide for every point in the x-y-plane whether it will be in $A_n$ for large n. ($A_n \subset A_{n+1}$)


*

*Any negative x: can never be inside the circle as every point with negative $x$ value is further from $(n,0)$ than $n$.

*Any point $(x,y)$ with positive x is included in a large enough circle that still touches $(0,0)$ and has its center on the x-axis.
The limit thus is $\{(x,y)|x\ge 0\}$.
A: Can you think of some points that belong to none of the circles, such as $(-1,-1)$? It should be pretty clear that a certain half-plane of points will not belong to any of the circles. (Can you prove this?) Consider points on the boundary of this half-plane, points in the interior of this half-plane, etc. and prove what seems evident from a picture. For some of these results, you can argue by making use of the Pythagorean Theorem applied to right triangles with a vertex at the origin and legs along the positive $x$- and $y$-axes. For other results, sufficiently large values of $n$ will be useful. For example, can you come up with a value of $N$ such that $n>N$ implies the circle associated with $n$ contains the point $\left(10^{100},10^{1000}\right)$?
