,
,
,
,
or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
,
isn't even rigorously defined in this expression. What exactly do we mean by 
.
,
does not exist in the first place, although you will see why in a calculus course. So the point is that 
?
indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that 
,
or ![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](http://latex.artofproblemsolving.com/2/6/0/260f97a1cf1ea8722ff97a29b83bac7bfe83e577.png)
etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function 
given by the mapping 
is undefined for 
.
is undefined because dividing by ![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](http://latex.artofproblemsolving.com/9/9/3/993c78b62ec2c84b5d77daa8e9a8cd6f70fcf904.png)
So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let 
for each 
which means that via this order topology each subset has an infimum and supremum and 
is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In ![\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}](http://latex.artofproblemsolving.com/6/d/8/6d8fc1eca5c791d430a7168cd3b37a02104fa318.png)
So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,![\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]](http://latex.artofproblemsolving.com/a/0/b/a0b67076efcc014c78b2023be1151c84b57c1503.png)
So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define 
as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by ![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](http://latex.artofproblemsolving.com/8/4/e/84ed3d6ab237df970333b87beaef5311caa46185.png)
which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, 
means complex infinity, since we are in the complex plane now. Here's the catch: division by ![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and 
Furthermore, we actually have some nice properties with multiplication that we didn't have before. In 
it holds that![\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]](http://latex.artofproblemsolving.com/8/e/0/8e035b4841d20c5ad671c46355cc667ca01e533b.png)
but 
and 
are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with ![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by 
that satisfy![\[f(xf(y) + f(x)) = f(x)f(y) + 2f(x) + f(y) - 1,\]](http://latex.artofproblemsolving.com/4/7/f/47f094ce57fe35d339ec3b93152ec3459a22ec48.png)
for every 
,
for every 
and 
,
.
be a set of 
points in a plane that lie within a disk of radius 
as picking a point 
and reflecting it across 
moves after which all points of 
Prove that
such that for every real number 
,
is a perfect square if and only if 
is a perfect square. Prove that 
is the square of a polynomial with real coefficients.
.
and 
intersect at the point 
.
?
,
.
are chosen from 0 to 1. What is the probability that their positive difference is more than 
?
?
families, there are 
children respectively. If a random child from any of the families is chosen, what is the probability that the child has 
siblings? Express your answer as a common fraction.[/quote]
,
people, exactly 
such that 
.
,
lies on diagonal 
such that 
.
.
has 
,
,
.
.
. If 
,
candies is 
times the probability of getting 
candies plus 
the probability of getting 
candies. What is the probability that Spencer gets 
candies when he plays the game? Express your answer as a common fraction
diverges for 
(harmonic sequence) but converges for 
because 
,
converges
Y by Davi-8191, OlympusHero, FaThEr-SqUiRrEl, lc426, centslordm, Adventure10, Designerd, jhu08, megarnie, Numbertheorydog, HWenslawski, Kanimet0, ImSh95, Zhaom, Mango247, Rounak_iitr, Math_.only., ItsBesi, cubres
Two circles 
and 
intersect at two points 
and 
. Let 
be the line tangent to these circles at 
and 
, respectively, so that lies closer to than . Let 
be the line parallel to and passing through the point , with 
on and 
on . Lines 
and 
meet at 
; lines 
and meet at 
; lines 
and meet at 
. Show that 
.
and 
intersect at two points 
and 
. Let 
be the line tangent to these circles at 
and 
, respectively, so that lies closer to than . Let 
be the line parallel to and passing through the point , with 
on and 
on . Lines 
and 
meet at 
; lines 
and meet at 
; lines 
and meet at 
. Show that 
.
Hence KA=KB.
Hence 
cuts the tangent length segment AB at its midpoint C. Since 
,
are centrally similar with similarity center N. Hence, M is the midpoint of the segment PQ. The radii 
of the circles 
are isosceles and the angles 
are equal. Since the lines 
are parallel and 
,
are equal and similarly, the angles 
are also equal. Thus the diagonal AB of the quadrilateral AMBE bisects its opposite angles at the vertices A, B, which implies that this quadrilateral is a kite with AE = AM, BE = BM and and its diagonals 
are perpendicular to each other. Consequently, the lines 
are also perpendicular to each other. Since M is the midpoint of the segment PQ, the line EM is the perpendicular bisector of this segment. It follows that the triangle 
is isosceles with EP = EQ.
bisects 
.
is the radical axis of 
,
.
and 
be the projections of 
,
.
,
(as 
.
and 
.
be the point such that 
and 
.
.
bisects 
.
and 
is concyclic.
and it is known that 
is the midpint of 
.
is radical axis for 
circles. also we know that radical axis passes through the midpoints of comment tangents. Thus 
lies on the same line. 
is the midpoint of 
.
[*] the last equation follows from the fact that 
is cyclic.
,
,
and since 
,
.
.
,
.
and 
,
and 
and therefore proving 
is equivalent to proving 
since 
.
,
because 
.
and 
and therefore 
as desired. 
and 
are isosceles and 
so 
.
is the midpoint of 
meets 
are not at all collinear.
is harmonic!
,
.
is cyclic, so 
.
.
,
by ASA congruency. Then 
is a kite, so 
,
.
,
,
.
,
are collinear, and 
are collinear, a homothety centered at 
takes 
is isosceles, and thus, 
as the point where they intersect has equivalent powers to the respective circles. From this we discover that 
due to the homothety at 
It then suffices to show 
.
![\[\angle EAB = \angle ECM = \angle ACM = \angle BAM. \]](http://latex.artofproblemsolving.com/6/4/e/64ea7480f2012c4f9052db0044dc365941c75e1e.png)
Similarly by symmetry, we find that 
Then by SAS, we conclude that 
as was to be shown. 
is a kite, so that 
as 
.
and 
so 
and 
.
.
As 
Taking the homothety centered at 
Then 
and similarly 
so by ASA congruence 
,
,
,
,
are reflections about 
=
from PoP we know 
and 
which means 
and since 
is a trapezoid, 
and we're done. label 
we know from 
that 
and 
being tangent that 
and 
the above facts indicate the similarity of 
so 
=> 
then clearly 
Therefore 
by the parallel lines.
Therefore 
and 
Similarly 
and 
Therefore, 
are the circumcenters of 
respectively so by Thales' Theorem it follows that 
Hence 
and we are done. QED
,
Thus, 
and 
and 
hence 
is a kite 
implying 
.
,
implies 
.
and 
.
![\[\measuredangle EAB = \measuredangle ACM = \measuredangle ANM = \measuredangle BAM = \measuredangle CMA\]](http://latex.artofproblemsolving.com/3/3/c/33cb77c57e505ab008a08d9152c6c0025237e8b4.png)
![\[\measuredangle EBA = \measuredangle BDM = \measuredangle BNM = \measuredangle ABM = \measuredangle DMB\]](http://latex.artofproblemsolving.com/1/7/b/17b761977cdf9496d808500e94dc78cebace60fd.png)
, implying that 
,
are isosceles. Therefore,![\[AE = AM = AC\]](http://latex.artofproblemsolving.com/0/5/9/059eab3cda6c66376c2362600d95b1f8aa3473db.png)
so 
,
is the cetner of 
.
.
,![\[LA = LB\]](http://latex.artofproblemsolving.com/4/a/1/4a15f4404bb42649f5bc6cf5f3df3284f9e64144.png)
and since 
,
,
,
.
,
,
bisects 
,
,
and similarly, 
.
.
,
is the perpendicular bisector of 
.
.
and 
,
.
and 
are isosceles. Furthermore, 
.
,
.
and similarly 
so 
and we're done
.
be the center of 
is the side bisector of 
,
.
and similarly 
,
(because they also share 
and analogously 
,
,
and similarly 
,
,
![[asy]unitsize(1.5cm);
import olympiad;
real root3 = sqrt(3);
pair O1 = (-root3, -1);
pair O2 = (root3, -1);
real radius = 2;
pair A = (-root3, 1);
pair B = (root3, 1);
pair M = (0, 0);
pair N = (0, -2);
pair C = (-2 * root3, 0);
pair D = (2 * root3, 0);
pair E = extension(A, C, B, D);
pair K = extension(A, B, M, N);
pair P = extension(A, N, C, D);
pair Q = extension(B, N, C, D);
draw(circle(O1, radius), heavycyan + dashed + linewidth(1.4));
draw(circle(O2, radius), heavymagenta + dashed + linewidth(1.4));
draw(A--B, heavygreen + linewidth(1.5));
draw(C--D, deepmagenta + linewidth(1.4));
draw(A--C, orange + linewidth(1.1));
draw(B--D, orange + linewidth(1.1));
draw(A--N, heavyblue + dashed + linewidth(1.1));
draw(B--N, heavyred + dashed + linewidth(1.1));
draw(E--N, black + dotted + linewidth(1.2));
draw(E--A, red + linewidth(2.5));
draw(E--B, blue + linewidth(2.5));
draw(E--P, cyan + linewidth(2.5));
draw(E--Q, magenta + linewidth(2.5));
dot(E, deepgreen);
dot(K, purple);
dot(P, brown);
dot(Q, orange);
dot(M, blue);
dot(N, red);
label("$G_1$", O1, SW, fontsize(9));
label("$G_2$", O2, SE, fontsize(9));
label("$A$", A, NW);
label("$B$", B, NE);
label("$M$", M, NE);
label("$N$", N, S);
label("$C$", C, W);
label("$D$", D, E);
label("$E$", E, NE);
label("$P$", P, SW);
label("$Q$", Q, S);
label("$K$", K, dir(45));
size(10cm);
pair min = (-2.5 * root3, -3.5);
pair max = (2.5 * root3, 3.5);
clip(box(min, max));
[/asy]](http://latex.artofproblemsolving.com/7/4/0/740fd20749ff72e5436121d9a8aa542c708ea487.png)
Let point 
Since 
lie on the same line, we can immediately see that this implies (from homothety) that 
.
.
All that remains to show now is that
(from 
)
which had to be proven. Therefore, we are done.
and 
and 
then by PoP, 
is the midpoint of 
by playing with Thales theorem on 
and 
