,
,
,
,
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 
![$P \in \mathbb R[x,y]$](http://latex.artofproblemsolving.com/8/a/0/8a0576299e976c51607959fb68c7c742b2a49c6a.png)
with ![$P \not\in \mathbb R[x] \cup \mathbb R[y]$](http://latex.artofproblemsolving.com/a/f/7/af71ba8c643d5bd8a2928fa6714ccd91713bd641.png)
and 
homeomorphismes of 
,
is an homoemorphisme too.
,
and 
are both integers.
be real numbers such that 
Prove that 
Let 
Prove that 
MathJuice bottles and Bob starts with 
MathJuice bottles. An unfair coin is then flipped, with probability 
of landing heads. If the coin lands heads, Alice gives Bob a bottle; otherwise, Bob gives Alice a bottle. This process repeats until someone runs out of bottles.
rest on a plane. A sphere with radius 
is tangent to all of them, but does not intersect nor lie on the plane. A sphere with radius 
lies on the plane, and is tangent to all three spheres with radius 
Compute the shortest distance between the sphere with radius 
Y by GeoKing
Let 
be a triangle with 
and let 
be the other intersection point of the angle bisector of 
with the circumcircle of the triangle . Let 
and 
be points on the sides 
and 
respectively, such that 
and let 
be the point of intersection of 
and 
. Let 
be the midpoint of 
. Prove that 
and the circumcircles of the triangles 
and 
pass through a common point.
be a triangle with 
and let 
be the other intersection point of the angle bisector of 
with the circumcircle of the triangle . Let 
and 
be points on the sides 
and 
respectively, such that 
and let 
be the point of intersection of 
and 
. Let 
be the midpoint of 
. Prove that 
and the circumcircles of the triangles 
and 
pass through a common point.
meet 
again at 
.
on 
be 
.
is simson line of 
.
and 
are homothetic . So 
.
with radius 
maps 
and hence 
i.e. line 
.
is mapped to itself. So 
.
denotes the image of 
under this inversion we get 
.
so we have (since P is the midpoint of EF)
so 
so 
so 
is cyclic
.
so 
and we are done
cut 
at 
.
cut 
.
midpoint arc 
of 
reflects 
.
lie on a circle.
and 
respectively.
intersects 
,
and 
and so 
is cyclic. Moreoever, 
and 
are cyclic too, since 
and 
.
is cyclic, as desired.
is the mid-point of 
are cyclic
are collinear
be the second intersections of the circumcircle of 
with 
and 
respectively. Let 
be the second intersections of the circumcircle of 
with 
and 
,
.
is cyclic and we're done as the desired intersection point is 
;
.
is cyclic with diameter 
.
are collinear since by spiral similarity and Menelaus we have 
.
is diameter of 
,
.
is cyclic, which will finish. Indeed, by spiral similarity again 
and 
)
.
,
is the antipode of 
,
are collinear, so we're done. 
be the feet of perpendicular from 
.
,
are collinear (Simson Line). Moreover, 
are cyclic. So, 
Consider an inversion with center 
.
,
,
.
be the intersection of 
so 
under inversion. Also, line 
.
,
and 
.
.
be the minor arc midpoint of 
(not containing 
(all these point must lie on the angle bisector of 
)
.![\[\measuredangle DPS = \measuredangle M_APS = \measuredangle DMS\]](http://latex.artofproblemsolving.com/1/9/7/197df6cfe1e87f903d400e478620dff98a43bc4c.png)
Thus, 
.
Further, let 
.
Now, one can note that clearly 
(angles of isosceles triangles with vertex 
Thus, the only possibility is that 
and thus, 
to be the intersection of circle 
.
and circle 
as the second intersection of circle 
are colinear since 
°
is cyclic.
.
is cyclic.
.
,
be the intersection of 
are collinear.
is concyclic.
,
.
cyclic. Let 
.
.
and 
are cyclic. By angle chase 
.
.
and we are done.
